{"id":281,"date":"2024-04-02T07:22:10","date_gmt":"2024-04-02T06:22:10","guid":{"rendered":"https:\/\/tutors4you.com\/?page_id=281"},"modified":"2024-04-02T07:22:57","modified_gmt":"2024-04-02T06:22:57","slug":"probability-tutorial","status":"publish","type":"page","link":"https:\/\/tutors4you.com\/index.php\/probability-tutorial\/","title":{"rendered":"Probability: basic concepts"},"content":{"rendered":"\n<div class=\"well well-sm\">\n\t\t\t\n\t\t\n        <h3>Random Experiment<\/h3> \n        <p>An experiment is said to be a random experiment, if it&#8217;s out-come can&#8217;t be\n        predicted with certainty.<\/p>\n        \n        <p><h4>Example<\/h4> If a coin is tossed, we can&#8217;t say, whether head or tail will appear. So it is a\n        random experiment.<\/p>\n        <h3>Sample Space<\/h3>\n         <p>The set of all possible out-comes of an experiment is called the sample\n        space. It is denoted by &#8216;S&#8217; and its number of elements\n        are n(s).<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice, the number that appears at top is any one of 1,2,3,4,5,6. So\n        here:<\/p>\n        \n        <p>S ={1,2,3,4,5,6} and n(s) = 6<\/p>\n        \n        <p>Similarly in the case of a coin, S={Head,Tail} or {H,T} and n(s)=2.<\/p>\n        \n        <h3>Elements<\/h3>\n\t\t<p>The elements of the sample space are called sample point or event point.<\/p>\n        <h3>Event<\/h3>\n        <p>Every subset of a sample space is an event. It is denoted by &#8216;E&#8217;.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice S={1,2,3,4,5,6}, the appearance of an event number will\n        be the event E={2,4,6}.<\/p>\n        \n        <p>Clearly E is a sub set of S.<\/p>\n        <h3>Simple event<\/h3>\n         <p>An event, consisting of a single sample point is called a simple event.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice, S={1,2,3,4,5,6}, so each of {1},{2},{3},{4},{5} and {6}\n        are simple events.<\/p>\n        <h3>Compound event<\/h3>\n        <p>A subset of the sample space, which has more than on element is called a mixed event.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice, the event of appearing of odd numbers is a compound event,\n        because E={1,3,5} which has &#8216;3&#8217; elements.<\/p>\n        <h3>Equally likely events<\/h3>\n         <p>Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other.<\/p>\n        \n        <p><h4>Example<\/h4> When a dice is thrown, all the six faces {1,2,3,4,5,6} are equally likely to come up.<\/p>\n        <h3>Exhaustive events<\/h3> \n         <p>When every possible out come of an experiment is considered.<\/p>\n        \n        <p><h4>Example<\/h4> A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of events.<\/p>\n        <h3>Classical definition of probability:<\/h3> \n        \n        <p>If &#8216;S&#8217; be the sample space, then the probability of occurrence of an event &#8216;E&#8217; is defined\n        as:<\/p>\n        \n        <\/br><\/br><img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/classical1.png\" class=\"img-responsive\">\n        \n        <p><h4>Example<\/h4> Find the probability of getting a tail in tossing of a coin.<\/p>\n       \n        <h4>Solution<\/h4>\n       \n        <p>Sample space S = {H,T} and\n        n(s) = 2<\/p>\n       \n        <p>Event &#8216;E&#8217; = {T} and n(E) = 1<\/p>\n       \n        <p>therefore  <\/br><\/br><img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/classical2.png\" class=\"img-responsive\"><\/p>\n        \n        <p>Note: This definition is not true, if<br>\n        (a) The events are not equally likely.<br>\n        (b) The possible outcomes are infinite.<\/p>\n        <h3>Sure event<\/h3>\n         <p>Let &#8216;S&#8217; be a sample space. If E is a subset of or equal to S then E is called a sure event.<\/p>\n        \n        <p><h4>Example<\/h4> In a throw of a dice, S={1,2,3,4,5,6}<\/p>\n        \n        <p>Let E<sub>1<\/sub>=Event of getting a number less than &#8216;7&#8217;.<\/p>\n        \n        <p>So &#8216;E<sub>1<\/sub>&#8216; is a sure event.<\/p>\n        \n        <p>So, we can say, in a sure event n(E) = n(S)<\/p>\n        <h3>Mutually exclusive or disjoint event<\/h3> \n        <p>If two or more events can&#8217;t occur simultaneously, that is no two of them can occur together.<\/p>\n        \n        <p><h4>Example<\/h4> When a coin is tossed, the event of occurrence of a head and the event of\n        occurrence of a tail are mutually exclusive events.<\/p>\n\t\t\n\t\t\t\t\n\t\t\t\n\t\t\t\t\n\t\t\t\t<p>Pictorial Representation:<\/p>\n\t\t\t\t<p><img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/mutuallyexclusiveevents.jpg\" class=\"img-responsive\"><\/p>\n\t\t\t\t<img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/classical4.png\"><\/p>\n\t\t\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n        <h3>Independent or mutually independent events<\/h3> \n        <p>Two or more events are said to be\n        independent if occurrence or non-occurrence of any of\n        them does not affect the probability of occurrence or\n        non-occurrence of the other event.<\/p>\n        \n        <p><h4>Example<\/h4> When a coin is tossed\n        twice, the event of occurrence of head in the first throw\n        and the event of occurrence of head in the second throw\n        are independent events.<\/p>\n        <h3>Difference between mutually exclusive an mutually independent events<\/h3>\n         <p>Mutually exclusiveness id used when the events are taken from the\n        same experiment, where as independence is used when the\n        events are taken from different experiments.<\/p>\n        <h3>Complement of an event<\/h3> \n         <p>Let &#8216;S&#8217; be the sample for random experiment, and &#8216;E&#8217; be an event, then\n        complement of &#8216;E&#8217; is  is denoted by E<sup>1<\/sup>.\n        Here E occurs, if and only if E<sup>1<\/sup> doesn&#8217;t occur.<\/p>\n        \n\t\t\t\t\n\t\t\t  <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/complementofanevent.jpg\" class=\"img-responsive\"><\/p>\n\t\t\t\t<p>Clearly <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/classical5.png\"><\/p>\n        \n        <h3>Random Experiment<\/h3>\n        <p> An experiment is said to be a random\n        experiment, if its out-come can&#8217;t be predicted with\n        certainly.<\/p>\n        \n        <p><h4>Example<\/h4> If a coin is tossed, we can&#8217;t\n        say, whether head or rail will appear. So it is a random\n        experiment.<\/p>\n        \n        <h3>Sample Space<\/h3> \n        <p>The set of all possible out-comes of an\n        experiment is called the sample &#8211; space.<\/p>\n        \n        <p>If a dice is thrown, the number, that\n        appears at top is any one of 1, 2, 3, 4, 5, 6,<\/p>\n        \n        <p>So here :<\/p>\n        \n        \n        <p>S = { 1, 2, 3, 4, 5, 6, } and n(s) = 6<\/p>\n        \n        <p>Similarly in the case of a coin, s = {H,T} and n (s) = 2.<\/p>\n        \n        <p>The elements of the sample of the\n        sample-space are called sample points or event points.<\/p>\n        \n        <p><h4>Example<\/h4> if S = {H, T}, than &#8216;H&#8217;\n        and &#8216;T&#8217; are sample points.<\/p>\n        \n        <p>Event:  Every\n        subset of a sample space is an event. It is denoted by\n        &#8216;E&#8217;.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice S = {1, 2, 3, 4, 5, 6,}, the appearance of an even number will be the\n        event E = {2, 4, 6}.<\/p>\n        \n       <p>Clearly <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/classical3.png\"><\/p>\n        \n        <h2>Important types of Events<\/h2>\n        \n        <h3>Simple or elementary event<\/h3>\n        <p>An event, consisting of a single point is\n        called a simple event.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice s = {1, 2,\n        3, 4, 5, 6} so each of {1}, {2}, {3}, {4}, {5} and {6} is\n        a simple event.<\/p>\n        \n        <h3>Compound or mixed event<\/h3> \n        <p>A subset of the sample space which has\n        more than one element is called a mixed event.<\/p>\n        \n        <p><h4>Example<\/h4> In throwing a dice, the event of odd\n        numbers appearing is a mixed event, because E = {1,\n        3, 5}, which has &#8216;3&#8217; elements.<\/p>\n        \n        <h3>Equally likely events<\/h3>\n        <p>Events are said to be equally likely, if\n        we have no reason to believe that one is more likely to\n        occur than the other.<\/p>\n        \n        <p><h4>Example<\/h4> When a dice is thrown, all the\n        six-faces {1, 2, 3, 4, 5, 6,} are equally likely to\n        come-up.<\/p>\n        \n        <h3>Exhaustive events<\/h3>\n        <p>When every possible outcome of an\n        experiment is considered, the observation is called\n        exhaustive events.<\/p>\n        \n        <p><h4>Example<\/h4> When a dice is thrown, cases 1, 2,\n        3, 4, 5, 5 form an exhaustive set of events.<\/p>\n        \n        \n\t\t\n\t\t\n\t\t<script async src=\"\/\/pagead2.googlesyndication.com\/pagead\/js\/adsbygoogle.js\"><\/script>\n\t\t<!-- t4uMediumRectangle -->\n\t\t<ins class=\"adsbygoogle\"\n\t\t\t style=\"display:inline-block;width:300px;height:250px\"\n\t\t\t data-ad-client=\"ca-pub-0715489232639450\"\n\t\t\t data-ad-slot=\"8639264715\"><\/ins>\n\t\t<script>\n\t\t(adsbygoogle = window.adsbygoogle || []).push({});\n\t\t<\/script>\n\t\t\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n        \n        \n        \n        <h2>Algebra of Events<\/h2><p> In a random\n        experiment, let &#8216;S&#8217; be the sample &#8211; space.<\/p>\n        \n        <p>Let A <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/subsetequalblue.jpg\" > S and B <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/subsetequalblue.jpg\" width=\"16\" height=\"12\"> S,\n        where &#8216;A&#8217; and &#8216;B&#8217; are events.<\/p>\n        \n        <p>Thus we say that :<\/p>\n       \n        <p>(i)\n        (A<img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" > B), is\n        an event occurs only when at least of &#8216;A&#8217;\n        and &#8216;B&#8217; occurs. <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\"\n        width=\"14\" height=\"12\"> (A <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\" height=\"12\">B) means (A or\n        B).<\/p>\n\t\t\n        <p >\n        <p><h4>Example<\/h4> if A =\n        { 2,4,6,} and B = {1, 6}, than the event &#8216;A&#8217; or\n        &#8216;B&#8217; occurs, if &#8216;A&#8217; or &#8216;B&#8217;\n        or both occur i.e. at least one of &#8216;A&#8217; and\n        &#8216;B&#8217; occurs. Clearly &#8216;A&#8217; or\n        &#8216;B&#8217; occur, if the out come is any one of the\n        outcomes 1, 2, 4, 6. That is A <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\"\n        width=\"10\" height=\"12\"> B. (From set &#8211; theory).<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        (A <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\"> B)\n        is an event, that occurs only when each one of\n        &#8216;A&#8217; and &#8216;B&#8217; occur <img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" >(A <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B)\n        means ( A and B).<\/p>\n        <p >\n        <p><h4>Example<\/h4> In the\n        above example, if the out come of an experiment is\n        &#8216;6&#8217;, then events &#8216;A&#8217; and\n        &#8216;B&#8217; both occur, because &#8216;6&#8217; is in\n        both sets. That is A <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\"\n        width=\"10\" height=\"12\"> B.<\/p>\n        <p >\n\t\t<br>\n        (iii)\n        A is an event, that occurs only when &#8216;A&#8217;\n        doesn&#8217;t occur.<\/p>\n\t\t<br>\n\t\t<br>\n\n\t\t<h2>Categories of Problems Related to\n        Probability <\/h2>\n        \n        <p><h4>Category A <\/h4> When n(E) and n(S) are determined by\n        writing down the elements of &#8216;E&#8217; and\n        &#8216;S&#8217;.<\/p>\n        \n        <p><h4>Category B <\/h4>When n(E) and n(S) are calculated by\n        the use of concept of permutation and combination.<\/p>\n        \n        <p><h4>Category C <\/h4> Problems based on P(E) + P(E<sup>1<\/sup>)\n        = 1<\/p>\n\t\t<br>\n\t\t<h3>Category A: When n(E) and n(S) are determined by\n        writing down the elements of &#8216;E&#8217; and &#8216;S&#8217;<\/h3>\n        \n       <p><h4>Question<\/h4> A coin\n        is tossed successively three times. Find the probability\n        of getting exactly one head or two heads.<\/p>\n        \n        <p><h4>Solution<\/h4> Let &#8216;S&#8217;\n        be the sample &#8211; space. Then,<\/p>\n        \n       \n        S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}<\/p>\n        \n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> n (S)\n        = 8<\/p>\n        <br>\n        Let &#8216;E&#8217; be the event of getting\n        exactly one head or two heads.<\/p>\n        \n        Then:<\/p>\n        \n       \n        E = { HHT, HTH, THH, TTH, THT, HTT }<\/p>\n        \n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> n\n        (E) = 6<\/p>\n        \n        Therefore:<\/p>\n        \n       \n        P(E) = n (E)\/ n (S)\n        = 6 \/\n        8\n        = 3\n        \/ 4<\/p>\n       <br>\n        <p><h4>Question<\/h4> Three\n        coins are tossed. What is the probability of getting (i)\n        all heads, (ii) two heads, (iii) at least one head, (iv)\n        at least two heads?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Let &#8216;S&#8217;\n        be the sample &#8211; space. Then<\/p>\n        <p >\n\t\t\n        S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }<\/p>\n        <p >\n\t\t<br>\n        (i)\n        Let &#8216;E<sub>1<\/sub>&#8216; = Event of getting all\n        heads.<\/p>\n        <p >\n       \n        Then E<sub>1<\/sub> = { HHH }<\/p>\n        <p >\n       \n        n (E<sub>1<\/sub>) = 1<\/p>\n        <p >\n       <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> P(E<sub>1<\/sub>)\n        = n (E<sub>1<\/sub>) \/ n(S)\n        = 1 \/ 8<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        Let E<sub>2<\/sub> = Event of getting\n        &#8216;2&#8217; heads.<\/p>\n        <p >\n       \n        Then:<\/p>\n        <p >\n       \n        E<sub>2<\/sub> = { HHT, HTH, THH }<\/p>\n        <p >\n       \n        n(E<sub>2<\/sub>) = 3<\/p>\n        <p >\n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> P (E<sub>2<\/sub>)\n        = 3 \/ 8<\/p>\n        <p >\n\t\t<br>\n        (iii)\n        Let E<sub>3<\/sub> = Event of getting at least one\n        head.<\/p>\n        <p >\n       \n        Then:<\/p>\n        <p >\n       \n        E<sub>3 <\/sub>= { HHH, HHT, HTH, THH, HTT,\n        THT, TTH }<\/p>\n        <p >\n       \n        n(E<sub>3<\/sub>) = 7<\/p>\n        <p >\n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\">\n        P (E<sub>3<\/sub>) = 7\n        \/ 8<\/p>\n        <p >\n\t\t<br>\n        (iv)\n        Let E<sub>4<\/sub> = Event of getting at least one\n        head.<\/p>\n        <p >\n       \n        Then:<\/p>\n        <p >\n       \n        E<sub>4 <\/sub>= { HHH, HHT, HTH, THH, }<\/p>\n        <p >\n       \n        n(E<sub>4<\/sub>) = 4<\/p>\n        <p >\n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\">\n        P (E<sub>4<\/sub>) =\n        4\/8\n       = 1\/2<\/p>\n        <p >\n\t\t<br>\n        <p><h4>Question<\/h4> What is\n        the probability, that a number selected from 1, 2, 3, &#8212;\n        2, 5, is a prime number, when each of the numbers is\n        equally likely to be selected.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> S = { 1, 2,\n        3, &#8212;- , 2, 5}\n        ><img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\">\n        n(S) = 25<\/p>\n        <p >\n        And E\n        = { 2, 3, 5, 7, 11, 13, 17, 19, 23 } <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> n(E)\n        = 9<\/p>\n        <p >\n        Hence P(E) = n(E)\n        \/ n(S)\n        = 9 \/ 25<\/p>\n      \n\t\t<br>\n        <p><h4>Question<\/h4> Two\n        dice are thrown simultaneously. Find the probability of\n        getting :<\/p>\n        \n        (i)\n        The same number on both dice,<\/p>\n        \n        (ii)\n        An even number as the sum,<\/p>\n        \n        (iii)\n        A prime number as the sum,<\/p>\n        \n        (iv)\n        A multiple of &#8216;3&#8217; as the sum,<\/p>\n        \n        (v)\n        A total of at least 0,<\/p>\n        \n        (vi)\n        A doublet of even numbers,<\/p>\n        \n        (vii)\n        A multiple of &#8216;2&#8217; on one dice and a multiple of\n        &#8216;3&#8217; on the other dice.<\/p>\n        \n        <p><h4>Solution<\/h4> Here:<\/p>\n        <p >\n       \n        S = { (1,1), (1,2) &#8212;&#8212;, (1,6), (2,1),\n        (2,2), &#8212;- (2,6), (3,1), (3,2), &#8212;&#8211;, (3,6),\n        (4,1), (4,2), &#8212;&#8212;&#8211; (4,6), (5,1), (5,2), &#8212;&#8211; (5,6),\n        (6,1,), (6,2), &#8212;&#8212;&#8212;&#8212;&#8212; (6,6) }<\/p>\n        \n       \n        n(S) = 6 x 6 = 36<\/p>\n        <br>\n        (i)\n        Let E<sub>1<\/sub> = Event of getting same number on both\n        side:<\/p>\n        \n       \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\" height=\"12\"> E<sub>1<\/sub>\n        = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }<\/p>\n        \n       \n        n(E<sub>1<\/sub>) =6<\/p>\n        \n       \n        P(E<sub>1<\/sub>) = n(E<sub>1<\/sub>)\/n(S)\n        = 6\/36\n        = 1\/6<\/p>\n        <br>\n        (ii)\n        Let E<sub>2<\/sub> = Event of getting an even number as\n        the sum.<\/p>\n        \n       \n        E<sub>2<\/sub> = { (1,1), (1,3), (1,5), (2,2), (2,4),\n        (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), <\/p>\n        <p >\n        (4,6), (5,1), (5,5), (6,2), (6,4), (6,6)\n        }<\/p>\n        \n       \n        n(E<sub>2<\/sub>) = 18 hence P(E<sub>2<\/sub>)\n        = n(E<sub>2<\/sub>)\/n(S)\n        =\n        18\/36\n        = 1\/2<\/p>\n        <br>\n        (iii)\n        Let E<sub>3<\/sub> = Event of getting a prime number as\n        the sum..<\/p>\n        \n       \n        E<sub>3<\/sub> = { (1,1), (1,2), (1,4), (1,6), (2,1),\n        (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), <\/p>\n        <p >\n        (5,2), (5,6), (6,1), (6,5),}<\/p>\n        \n        n(E<sub>3<\/sub>) = 15 <\/p>\n        \n        P(E<sub>2<\/sub>)\n        = n(E<sub>3<\/sub>) \/\n        n(S) =\n        15\/36\n        = 5\/12<\/p>\n        <br>\n        (iv)\n        Let E<sub>4<\/sub> = Event of getting a multiple of\n        &#8216;3&#8217; as the sum.<\/p>\n        \n       \n        E<sub>4<\/sub> = { (1,2), (1,5), (2,1), (2,4), (3,3),\n        (3,6), (4,2), (4,5), (5,1), (5,4), <\/p>\n        \n       \n        (6,3), (6,6),}<\/p>\n        \n       \n        n(E<sub>4<\/sub>) = 12 <\/p>\n        \n        P(E<sub>4<\/sub>)\n        = n(E<sub>4<\/sub>)\/n(S)\n        =\n        12\/36\n        = 1\/3<\/p>\n        <br>\n        (v)\n        Let E<sub>5<\/sub> = Event of getting a total of at least\n        10.<\/p>\n        \n       \n        E<sub>5<\/sub> = { (4,6), (5,5), (5,6), (6,4), (6,5),\n        (6,6), }<\/p>\n        \n       \n        n(E<sub>5<\/sub>) = 6 <\/p>\n        \n        P(E<sub>5<\/sub>)\n        = n(E<sub>5<\/sub>)\/n(S)\n        = 6\/36\n        =\n        1\/6<\/p>\n        \n       \n        <\/p>\n        <br>\n        (vi)\n        Let E<sub>6<\/sub> = Event of getting a doublet of even\n        numbers.<\/p>\n        \n       \n        E<sub>6<\/sub> = { (2,2), (4,4), (6,6), }<\/p>\n        \n       \n        n(E<sub>6<\/sub>) = 3 <\/p>\n        \n        P(E<sub>6<\/sub>)\n        = n(E<sub>6<\/sub>)\/\n        n(S) =\n        3\/36\n        =\n        1\/12<\/p>\n       <br>\n\t   <p >\n        (vii) Let E<sub>7<\/sub>\n        = Even of getting a multiple of &#8216;2&#8221; on one dice\n        and a multiple of &#8216;3&#8217; on the other dice.<\/p>\n        \n       \n        E<sub>7<\/sub> = { (2,3), (2,6), (4,3), (4,6), (6,3),\n        (3,2), (3,4), (3,6), (6,2), (6,4) }<\/p>\n        \n       \n        n(E<sub>7<\/sub>) = 11<\/p>\n        \n        P(E<sub>7<\/sub>)\n        = n(E<sub>7<\/sub>) \/\n        n(S) = 11\/36<\/p>\n        <br>\n        <p><h4>Question<\/h4> What is the\n        probability, that a leap year selected at random will\n        contain 53 Sundays?<\/p>\n        \n        <p><h4>Solution<\/h4> A leap year has\n        366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.<\/p>\n        \n       \n        <p>The remaining 2 days may be any of the following :<\/p>\n        \n        <p>(i)\n        Sunday and Monday<\/p>\n        \n        <p>(ii)\n        Monday and Tuesday<\/p>\n        \n        <p>(iii)\n        Tuesday and Wednesday<\/p>\n        \n        <p>(iv)\n        Wednesday and Thursday<\/p>\n        \n        <p>(v)\n        Thursday and Friday<\/p>\n        \n        <p>(vi)\n        Friday and Saturday<\/p>\n        \n        <p>(vii)\n        Saturday and Sunday<\/p>\n        \n        <p>For having 53 Sundays in a year, one of\n        the remaining 2 days must be a Sunday.<\/p>\n        \n       \n        <p>n(S) = 7<\/p>\n        \n       \n        <p>n(E) = 2<\/p>\n        \n       \n        <p>P(E) = n(E) \/ n(S)\n        = 2 \/ 7<\/p>\n\t\t\n\t\t<h3>Category B: Problems based on fundamental principal\n        of counting and permutations and combinations<\/h3>\n       \n\t\t<br>\n        <p><h4>Question<\/h4> A bag\n        contains &#8216;6&#8217; red, 4 white and 8 blue balls. If\n        three balls are drawn at random, find the probability,\n        that<\/p>\n        <p >\n        (i)\n        &#8216;1&#8217; is red and &#8216;2&#8217; are white, (ii)\n        &#8216;2&#8217; are blue and 1 is red, (iii) none is red.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We have to select\n        &#8216;3&#8217; balls, from 18 balls (6+4+8)<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = <sup>18<\/sup>C<sub>3<\/sub> = 18! \/ (3!\n        x 15!) = (18x17x16) \/ (3x2x1) = 816<\/p>\n        <p >\n\t\t<br>\n        (i)\n        Let E<sub>1<\/sub> = Event of getting &#8216;1&#8217; ball\n        is red and &#8216;2&#8217; are white<\/p>\n        <p >\n        Total number of ways = n(E<sub>1<\/sub>)\n        = <sup>6<\/sup>C<sub>1<\/sub> x <sup>4<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        = 6! \/ (1! x 5!) x 4! \/ (2! x 2!)<\/p>\n        <p >\n        = 6 x 4 \/ 2<\/p>\n        <p >\n        = 36<\/p>\n        <p >\n        P (E<sub>1<\/sub>) = n (E<sub>1<\/sub>) \/\n        n(S) = 36\/816 = 3\/68<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        Let E<sub>2<\/sub> = Event of getting &#8216;2&#8217; balls\n        are blue and &#8216;1&#8217; is red.<\/p>\n        <p >\n        = Total no. of ways <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\"\n        width=\"13\" height=\"12\"> n(E<sub>2<\/sub>) = <sup>8<\/sup>C<sub>2<\/sub>\n        x <sup>6<\/sup>C<sub>1<\/sub><\/p>\n        <p >\n        = 8! \/ (2! x 6!) x 6! \/ (1! x\n        5!)<\/p>\n        <p >\n        = (8 x 7) \/ 2 x 6 \/ 1 = 168<\/p>\n        <p >\n        P(E<sub>2<\/sub>) = 168 \/\n        816 = 7\/34<\/p>\n        <p >\n\t\t<br>\n        (iii)\n        Let E<sub>3 <\/sub>= Event of getting &#8216;3&#8217;\n        non &#8211; red balls. So now we have to choose all the\n        three balls from 4 white and 8 blue balls.<\/p>\n        <p >\n        Total number of ways :<\/p>\n        <p >\n        n(E<sub>3<\/sub>) = <sup>12<\/sup>C<sub>3<\/sub>\n        = 12! \/ (3! x 9!) = (12x11x10) \/\n        (3x2x1) = 220<\/p>\n        <p >\n        P(E<sub>3<\/sub>) = n(E<sub>3<\/sub>) \/\n        n(S) = 220 \/ 816 = 55\/204<\/p>\n        <p >\n\t\t<br>\n        <p><h4>Question<\/h4>\n        A box contains 12 bulbs of which &#8216;4&#8217; are\n        defective. All bulbs took alike. Three bulbs are drawn\n        randomly.<\/p>\n        <p >\n        What is the probability that :<\/p>\n        <p >\n        (i)\n        all the &#8216;3&#8217; bulbs are defective?<\/p>\n        <p >\n        (ii)\n        At least &#8216;2&#8217; of the bulbs chosen are defective?<\/p>\n        <p >\n        (iii)\n        At most &#8216;2&#8217; of the bulbs chosen are defective?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We have to select\n        &#8216;3&#8217; bulbs from 12 bulbs.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = <sup>12<\/sup>C<sub>3<\/sub>\n        = 12! \/ (3! x 9!) = (12x11x10) \/\n        (3x2x1) = 220<\/p>\n        <p >\n\t\t<br>\n        (i)\n        Let E<sub>1<\/sub> = All the &#8216;3&#8217;\n        bulbs are defective.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> All bulbs have been chosen, from\n        &#8216;4&#8217; defective bulbs.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(E<sub>1<\/sub>) = <sup>4<\/sup>C<sub>3<\/sub>\n        = 4! \/ (3! x 1!) = 4<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>1<\/sub>) = n(E<sub>1<\/sub>) \/\n        n(S) = 4 \/220 = 1\/55<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        Let E<sub>2<\/sub> = Event drawing at least 2\n        defective bulbs. So here, we can get &#8216;2&#8217;\n        defective and 1 non-defective bulbs or 3 defective bulbs.<\/p>\n        <p >\n        n(E<sub>2<\/sub>) = <sup>4<\/sup>C<sub>2<\/sub>\n        x <sup>8<\/sup>C<sub>1<\/sub> + <sup>4<\/sup>C<sub>3<\/sub>\n        [Non-defective bulbs = 8]<\/p>\n        <p >\n        = 4! \/ (2! x 2!) x 8! \/ (1! x 7!) + 4! \/\n        (3! x 1!)<\/p>\n        <p >\n        = 4&#215;3 \/ 2 x 8\/1 + 4\/1 = 48+4<\/p>\n        <p >\n        n(E<sub>2<\/sub>) = 52<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>2<\/sub>) = n(E<sub>2<\/sub>) \/ n(S) =\n        52\/220 = 13\/55<\/p>\n        <p >\n\t\t<br>\n        (iii)\n        Let E<sub>3<\/sub> = Event of drawing at most\n        &#8216;2&#8217; defective bulbs. So here, we can get no\n        defective bulbs or 1 is defective and &#8216;2&#8217; is\n        non-defective or &#8216;2&#8217; defective bulbs.<\/p>\n        <p >\n        n(E<sub>3<\/sub>) = <sup>8<\/sup>C<sub>3<\/sub>\n        + <sup>4<\/sup>C<sub>1<\/sub> x <sup>8<\/sup>C<sub>2<\/sub> +\n        <sup>4<\/sup>C<sub>2<\/sub> x <sup>8<\/sup>C<sub>1<\/sub><\/p>\n        <p >\n        = 8? \/ (3? x 5?) + 4? \/ (1? x 3?) x 8? \/\n        (2? x 6?) + 4? \/ (2? x 2?) x 8? \/ (1? x 7?)<\/p>\n        <p >\n        = (8x7x6) \/ (3x2x1) + 4 x (8&#215;7)\/2 + (4 x\n        3) \/ 2 + 8\/1<\/p>\n        <p >\n        = 216<\/p>\n        <p >\n        P(E<sub>3<\/sub>) = n(E<sub>3<\/sub>) \/\n        n(S) = 216 \/ 220 = 54 \/ 55<\/p>\n        <p >\n\t\t<br>\n        <p><h4>Question<\/h4>\n        In a lottery of 50 tickets numbered from &#8216;1&#8217; to\n        &#8217;50&#8217; two tickets are drawn simultaneously. Find\n        the probability that:<\/p>\n        <p >\n\t\t\n        (i)\n        Both the tickets drawn have prime number on them,<\/p>\n        <p >\n\t\t\n        (ii)\n        None of the tickets drawn have a prime number on it.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We want to select &#8216;2&#8217;\n        tickets from 50 tickets.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\">n(S) = <sup>50<\/sup>C<sub>2<\/sub> = 50? \/ (2?\n        x 48?) = (50&#215;49) \/ 2 = 1225<\/p>\n        <p >\n\t\t<br>\n        (i)\n        Let E<sub>1<\/sub> = Event that both the tickets have\n        prime numbers Prime numbers between &#8216;1&#8217; to\n        &#8217;50&#8217; are :<\/p>\n        <p >\n        2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.<\/p>\n        <p >\n        Total Numbers = 15.<\/p>\n        <p >\n        We have to select &#8216;2&#8217; numbers\n        from these 15 numbers.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(E<sub>1<\/sub>) = <sup>15<\/sup>C<sub>2<\/sub>\n        = 15? \/ (2? x 13?) = (15&#215;47) \/ 2 = 105<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>1<\/sub>) = n(E<sub>1<\/sub>) \/ n(S) =\n        105\/1225 = 21\/245<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        Non prime numbers between &#8216;1&#8217; to &#8217;50&#8217;\n        = 50-15 = 35<\/p>\n        <p >\n        Let E<sub>2<\/sub> = Event that both the\n        tickets have non-prime numbers.<\/p>\n        <p >\n        Now we have to select &#8216;2&#8217;\n        numbers, from &#8217;35&#8217; numbers.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\">n(E<sub>2<\/sub>) = <sup>35<\/sup>C<sub>2<\/sub>\n        = 35? \/ (2? x 33?) = (35&#215;34) \/ 2 = 595<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>2<\/sub>) = n(E<sub>2<\/sub>) \/ n(S) =\n        595 \/ 1225 = 17\/35<\/p>\n        \n        <p><h4>Question<\/h4> A\n        bag contains 30 tickets, numbered from &#8216;1&#8217; to\n        &#8217;30&#8217;. Five tickets are drawn at random and\n        arranged in ascending order. Find the probability that\n        the third number is 20.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Total number of\n        ways of selecting &#8216;5&#8217; tickets from 30 tickets =\n        <sup>30<\/sup>C<sub>5<\/sub><\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = <sup>30<\/sup>C<sub>5<\/sub> = 30? \/\n        (5? x 25?) = (30 x 29 x 28 x 27 x 26) \/ (5 x 4 x 3 x 2 x\n        1)<\/p>\n        <p >\n        n(S) = 29 x 27 x 26 x 7<\/p>\n        <p >\n        Suppose the &#8216;5&#8217; tickets are a1,\n        a2,20, a4, a5<\/p>\n        <p >\n        They are arranged in ascending order.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> a1, a2 <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/subsetequalgreen.jpg\"\n        width=\"16\" height=\"12\"> {1, 2, 3, &#8212;&#8212;- , 19} and a4,\n        a5 <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/subsetequalgreen.jpg\" width=\"16\" height=\"12\">\n        { 21, 22, 23, &#8212;&#8211;, 30}<\/p>\n        <p >\n        We have to select &#8216;2&#8217; tickets\n        from first &#8217;19&#8217; tickets and &#8216;2&#8217;\n        tickets from last 10 tickets.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(E) = <sup>19<\/sup>C<sub>2<\/sub>\n        x <sup>10<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        = 19? \/ (2? x 17?) = 10? \/ (2? x 8?) =\n        (19 x 18) \/ 2 = (10 x 9) \/ 2<\/p>\n        <p >\n        = 19 x 9 x 5 x 9<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E) = n(E) \/ n(S) = (19x9x5x9) \/\n        (29x27x26x7) = 285 \/ 5278<\/p>\n        <h3>\n        Odds in Favour and Odds against an Event:<\/h3>\n        <p >\n        Let &#8216;S&#8217;; be the sample space and\n        &#8216;E&#8217; be an event. Let &#8216;E&#8217; devotes the\n        complement of event &#8216;E&#8217;, then.<\/p>\n        <p >\n\t\t<br>\n        (i)\n        Odds in favour of event &#8216;E&#8217; = n(E) \/ n(E<sup>1<\/sup>)<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        Odds in against of an event &#8216;E&#8217; = n(E<sup>1<\/sup>)\n        \/ n(E)<\/p>\n        <p >\n        Note : Odds in favour of &#8216;E&#8217; =\n        n(E) \/ n(E<sup>1<\/sup>)<\/p>\n        <p >\n        = [n(E) \/ n(S)] \/ [n(E<sup>1<\/sup>) \/\n        n(S)] = P(E) \/ P(E<sup>1<\/sup>)<\/p>\n        <p >\n        Similarly odds in against of\n        &#8216;E&#8217; = P(E<sup>1<\/sup>) \/ P(E)<\/p>\n        <p><h4>Example<\/h4> The odds in favours of an event are\n        3:5 find the probability of the occurrence of this event.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Let &#8216;E&#8217; be an\n        event.<\/p>\n        <p >\n        Then odds in favour of E = n(E) \/ n(E<sup>1<\/sup>)\n        = 3 \/ 5<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(E) = 3 and n(E<sup>1<\/sup>) = 5<\/p>\n        <p >\n        Total number of out-comes n(S) = n(E) +\n        n(E<sup>1<\/sup>) = 3+5 = 8<\/p>\n        <p >\n        P(E) = n(E) \/ n(S) = 3 \/ 8<\/p>\n        <p >\n        <p><h4>Question<\/h4>\n        If &#8217;12&#8217; persons are seated at a round table,\n        what is the probability that two particulars persons sit\n        together?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We have to arrange 12 persons along\n        a round table.<\/p>\n        <p >\n        So if &#8216;S&#8221; be the sample &#8211;\n        space, then n(S) = (12-1)? = 11?<\/p>\n        <p >\n        n(S) = 11?<\/p>\n        <p >\n        Now we have to arrange the persons in\n        away, such that &#8216;2&#8217; particulars person sit\n        together.<\/p>\n        <p >\n        Regarding that 2 persons as one person,\n        we have to arrange 11 persons.<\/p>\n        <p >\n        Total no. of ways = (11-1)? = 10? ways.<\/p>\n        <p >\n        That &#8216;2&#8217; persons can be\n        arranged among themselves in 2? ways.<\/p>\n        <p >\n        So, total no. of ways, of arranging 12\n        persons, along a round table, so that two particular\n        person sit together : = 10? x 2?<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\">n(E) = 10? x 2?<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E) = n(E) \/ n(S) = (10? x 2?) \/ 11? = 2 \/\n        11<\/p>\n        <p >\n        <p><h4>Question<\/h4> 6\n        boys and 6 girls sit in a row randomly, find the\n        probability that all the &#8216;6&#8217; girls sit\n        together.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We have to\n        arrange &#8216;6&#8217; boys and &#8216;6&#8217; girls in a\n        row.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = 12?<\/p>\n        <p >\n        Now, we have to arrange &#8216;6&#8217;\n        girls in a way, such that all of them should sit\n        together.<\/p>\n        <p >\n        Regarding all the 6 girls as one person,\n        we have to arrange 7 person in a row.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> Total no. of ways = 7?<\/p>\n        <p >\n        But 6 girls can be arranged among\n        themselves in 6? ways.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(E) = 7? x 6?<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E) = n(E) \/ n(S) = (7? x 6?) \/ 12? =\n        (6x5x4x3x2x1) \/ (12x11x10x9x8)<\/p>\n        <p >\n        P(E) = 1 \/ 132<\/p>\n        <p><h4>Question<\/h4>\n        If from a pack of &#8217;52&#8217; playing cards one card\n        is drawn at random, what is the probability that it is\n        either a kind or a queen?<\/p>\n        <p><h4>Solution<\/h4> n(S) = Total\n        number of ways of selecting 1 card out of 52 cards.<\/p>\n        <p >\n        = <sup>52<\/sup>C<sub>1<\/sub> = 52<\/p>\n        <p >\n        n(E) = Total number of selections of a\n        card, which is either a kind or a queen.<\/p>\n        <p >\n        = <sup>4<\/sup>C<sub>1<\/sub> + <sup>4<\/sup>C<sub>1<\/sub>\n        = 4 + 4 = 8<\/p>\n        <p >\n        P(E) = n(E) \/ n(S) = 8 \/ 52 = 2 \/ 3<\/p>\n        <p >\n        <p><h4>Question<\/h4>\n        From a pack of 52 playing cards, three cards are drawn at\n        random. Find the probability of drawing a king, a queen\n        and a jack.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Here n(S) = <sup>52<\/sup>C<sub>3<\/sub>\n        = 52? \/ (3? x 49?) = (52x51x50) \/ (3x2x1)<\/p>\n        <p >\n        = 52x17x25<\/p>\n        <p >\n        n(E) = <sup>4<\/sup>C<sub>1<\/sub>. <sup>4<\/sup>C<sub>1<\/sub>.\n        <sup>4<\/sup>C<sub>1<\/sub><\/p>\n        <p >\n        = 4? \/ (1? x 3?) x 4? \/ (1? x 3?) = 4? \/\n        (1? x 3?)<\/p>\n        <p >\n        n(E) = 4 x 4 x 4<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E) = n(E) \/ n(S) = (4x4x4) \/\n        (52x17x25) = 16 \/ 5525<\/p>\n        \n\t\t<h3>Category C:Problems based on finding P(E<sup>1<\/sup>),\n        by the use of P(E) + P(E<sup>1<\/sup>)\n        = 1 <\/h3>\n\t\t\n       <p> Note : When an event has a lot of out\n        comes, then we use this concept.<\/p>\n       \n        <p><h4>Question<\/h4> What\n        is the probability of getting a total of less than\n        &#8217;12&#8217; in the throw of two dice?<\/p>\n       \n        <p><h4>Solution<\/h4> Here n(S) = 6x6x\n        = 36<\/p>\n        <p >\n        It is very difficult to find out all the\n        cares, in which we can find the total less then\n        &#8217;12&#8217;.<\/p>\n        <p >\n        So let E = The event, that the sum of\n        numbers is &#8217;12&#8217;.<\/p>\n        <p >\n        Then E = { 6, 6}<\/p>\n        <p >\n        n(E) = 1<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E) = n(E) \/ n(S) = 1\/36<\/p>\n        <p >\n        Required probability, P(E<sup>1<\/sup>) =\n        1-P(E)<\/p>\n        <p >\n        = 1 &#8211; 1\/36<\/p>\n        <p >\n        P(E1) = 35 \/36<\/p>\n        <p><h4>Question<\/h4> There\n        are &#8216;4&#8217; envelopes corresponding to\n        &#8216;4&#8217; letters. If the letters are placed in the\n        envelopes at random, what is the probability that all the\n        letters are not placed in the right envelopes?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> We have to place\n        &#8216;4&#8217; letters in 4 envelopes.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = 4!<\/p>\n        <p >\n        Now:<\/p>\n        <p >\n        Let E = The event, that all the 4 letters\n        are placed in the corresponding envelopes.<\/p>\n        <p >\n        So E<sup>1<\/sup> = The event that all the\n        &#8216;4&#8217; letters are not placed in the right\n        envelope.<\/p>\n        <p >\n        Here n(E) = 1<\/p>\n        <p >\n        P(E) = n(E) \/ n(S)\n        = 1 \/ 4! = 1 \/ 24<\/p>\n        <p >\n        Required probability, P(E<sup>1<\/sup>)\n        = 1- P(E)<\/p>\n        <p >\n        = 1 &#8211; (1\/24)<\/p>\n        <p >\n        P(E<sup>1<\/sup>) = 23 \/ 24<\/p>\n        <p >\n\t\t<br>\n        <h2>Total Probability<\/h2>\n        \n        <p><h3>Theorem &#8211; 1<\/h3>  If &#8216;A&#8217; and\n        &#8216;B&#8217; are mutually exclusive events then P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B) = 0\n        or P ( A and B) = 0<\/p>\n        <p >\n        <p><h4>Proof<\/h4> If &#8216;A&#8217; and\n        &#8216;B&#8217; are mutually exclusive events then A <img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" class=\"img-responsive\"> B = <img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\" ><\/p>\n        \n        <p><img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\"> P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = P(<img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\" >)<\/p>\n        <p >\n        = n(<img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\">) \/ n(S) [ By definition of\n        probability]<\/p>\n        <p><img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/mutuallyexclusiveevents.jpg\" class=\"img-responsive\"><\/p>\n        <p >\n        <p>= 0 \/ n(S)\n        [Since the number of elements in a null &#8211; set is\n        &#8216;0&#8217;]<\/p>\n        <p >\n        <p>P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = 0<\/p>\n        <p >\n        <h3>Addition Theorem of Probability<\/h3>\n\t\t\n        <p> If &#8216;A&#8217; and &#8216;B&#8217;\n        by any two events, then the probability of occurrence of\n        at least one of the events &#8216;A&#8217; and\n        &#8216;B&#8217; is given by:<\/p>\n        <p >\n        P(A or B) = P(A) + P(B) &#8211; P (A and\n        B)<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B) = P(A) + P(B) &#8211; P (A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p ><img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/additiontheorem.jpg\" class=\"img-responsive\"><\/p>\n        <p >\n        From set theory, we have :<\/p>\n        <p >\n        n(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B) = n(A) + n(B) &#8211; n(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        Dividing both sides by n(S) :<\/p>\n        <p >\n        n(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B) \/ n(S) = n(A) \/ n(S)\n        + n(B) \/ n(S) &#8211; n(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B) \/\n        n(S)<\/p>\n        <p >\n        or P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">B) = p(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        Corollary : If &#8216;A&#8217; and\n        &#8216;B&#8217; are mutually exclusive events,<\/p>\n        <p >\n        Then P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = 0. [ As we have proved]<\/p>\n        <p >\n        In this case :<\/p>\n        <p class=\"MsoNormal\"\n        style=\"text-align: justify; text-indent: -.25in; margin-left: .5in\">\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\"> P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">B) = p(A) + P(B)<\/p>\n        <p >\n\t\t<h3>Addition theorem for &#8216;3&#8217; events\n        &#8216;A&#8217;, &#8216;B&#8217; and &#8216;C&#8217;<\/h3>\n        \n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">C) = P(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)\n        &#8211; P(B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C) &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">C) + P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C)<\/p>\n        <p><h4>Proof<\/h4> P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">C) = P[(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">B) <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">C]<\/p>\n        <p >\n        = P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">b) + P(C) &#8211; P[(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\">B) <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">C]\n       \n        [ By addition theorem for two events]<\/p>\n        <p >\n        = P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B) &#8211; P(C) &#8211; [P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C) +\n        P(B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C)\n        &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C)]<\/p>\n        <p >\n        = P(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B) +\n        P(C) &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C) &#8211; P(B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">C) + P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C)<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">C) = P(A) + P(B) + P(C) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)\n        &#8211; P(B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C) &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">C) + P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C)<\/p>\n        <p >\n        Corollary : If &#8216;A&#8217;,\n        &#8216;B&#8217; and &#8216;C&#8217; are mutually exclusive\n        events, then P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = 0, P(B<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C) = 0,\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">C)\n        = 0 and P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">C) = 0.<\/p>\n        <p >\n        In this case :<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\"> P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">B<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\">C) = P(A) + P(B) + P(C)<\/p>\n        <p >\n\t\t<h3>General Form of Addition Theorem of\n        Probability<\/h3>\n       \n        <p >\n  \n        n<\/p>\n        <p >\n        P(A<sub>1<\/sub> <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\"> A<sub>2<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\" height=\"12\"> &#8212;&#8211; <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\" height=\"12\"> A<sub>n<\/sub>)\n        = <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/sigmasingleblue.jpg\" width=\"19\"\n        height=\"24\">P(A<sub>i<\/sub>) &#8211; <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/sigmasingleblue.jpg\" width=\"19\" height=\"24\"> P(A<sub>i<\/sub>\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\"> A<sub>j<\/sub>)<\/p>\n        <p >\n       \n        i=1\n       i&lt;j<\/p>\n        <p >\n        + <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/sigmasingleblue.jpg\"\n        width=\"19\" height=\"24\">P(A<sub>i<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">A<sub>j<\/sub>\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\"> A<sub>k<\/sub>)\n        &#8212;&#8211; (-1)<sup>n-1<\/sup> P(A<sub>1<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\"> A<sub>2<\/sub>\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">\n        &#8212;- <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\"> A<sub>n<\/sub>)<\/p>\n        <p >\n         i&lt;j&lt;k<\/p>\n        <p >\n        Corollary : For any number of mutually\n        exclusive events, A<sub>1<\/sub>, A<sub>2<\/sub>, &#8212;&#8212; ,\n        A<sub>n<\/sub> :<\/p>\n        <p >\n        P(A<sub>1<\/sub> <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\"> A<sub>2<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\" height=\"12\"> &#8212;- <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\" height=\"12\">A<sub>n<\/sub>)\n        = P(A<sub>1<\/sub>) + P(A<sub>2<\/sub>) + &#8212;&#8212; +\n        P(A<sub>n<\/sub>)<\/p>\n        <p >\n\t\t<h3>Theorem &#8211; 3 : For any two events\n        &#8216;A&#8217; and &#8216;B&#8217;<\/h3>\n       \n        <p >\n        P(A-B) = P(A) &#8211; P(A<img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" class=\"img-responsive\">B) =\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B<sup>1<\/sup>)<\/p>\n        <p><img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/additiontheorem.jpg\" width=\"225\" height=\"134\"><\/p>\n        <p >\n        From the figure:<\/p>\n        <p >\n        (A-B) <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\"> (A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\"\n        > &#8212;&#8212;&#8212;&#8212;&#8212;&gt; (i)<\/p>\n        <p >\n        and <\/p>\n        <p class=\"MsoNormal\"\n        style=\"text-align:justify;text-indent:.5in\">\n        (A-B) <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\" width=\"10\"\n        height=\"12\"> (A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">B) = A<\/p>\n        <p >\n        P[(A-B) <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/unionblue.jpg\"\n        width=\"10\" height=\"12\"> (A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B)] = P(A)<\/p>\n        <p >\n        or P(A-B) + P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B) =\n        P(A) <\/p>\n        <p >\n        [From (i) (A-B)<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\"> (A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)\n        = <img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\" >\n        i.e. These events are mutually exclusive]<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\"> P(A-B) = P(A) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        or P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\">B) = P(A) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        Similarly P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B) =\n        P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">B)<\/p>\n        <p >\n        Proof of P(E) + P(E<sup>1<\/sup>) = 1, by\n        the addition theorem of probability:<\/p>\n        <p >\n        We know that :<\/p>\n        <p >\n        P(A <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\">\n        B) = P(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        Putting A = E and B = E<sup>1<\/sup><\/p>\n        <p >\n        P (E <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\">\n        E<sup>1<\/sup>) = P(E) + P(E<sup>1<\/sup>)\n        &#8211; P (E<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\" width=\"10\"\n        height=\"12\">E<sup>1<\/sup>) &#8212;&#8212;&#8212;&#8212;&#8212;-&gt; (1)<\/p>\n        <p >\n        From set theory : E \n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\"> > E<sup>1<\/sup>\n        = S<\/p>\n        <p >\n        And E <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectionblue.jpg\"\n        width=\"10\" height=\"12\"> E<sup>1<\/sup> = <img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\" ><\/p>\n        <p >\n        From:<\/p>\n        <p >\n        \n        P(S) = P(E) + P(E<sup>1<\/sup>) &#8211; P(<img decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\" >)<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrowblue.jpg\" width=\"14\"\n        height=\"12\"> 1 = P(E) + P(E<sup>1<\/sup>)\n        &#8211; 0<\/p>\n        <p >\n        or P(E) + P(E<sup>1<\/sup>) = 1<\/p>\n        <p >\n        [P(S) = 1, P(<img decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/phiblue.png\"\n        >) = 0]<\/p>\n        <p > <b><u>EXAMPLES<\/u><\/b><\/p>\n        <p >\n\t\t<h3>Problems based on addition theorem of\n        probability<\/h3>\n       \n        <p >\n        Working rule : <\/p>\n        <p >\n\t\t<br>\n        (i)\n        A <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\"> B\n        denotes the event of occurrence of at least one of the\n        event &#8216;A&#8217; or &#8216;B&#8217;<\/p>\n        <p >\n\t\t<br>\n        (ii)\n        A <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">\n        B denotes the event of occurrence of both the events\n        &#8216;A&#8217; and &#8216;B&#8217;.<\/p>\n        <p >\n\t\t<br>\n        (iii)\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\">B) or\n        P(A+B) denotes the probability of occurrence of at least\n        one of the event &#8216;A&#8217; or &#8216;B&#8217;.<\/p>\n        <p >\n\t\t<br>\n        (iv)\n        P(<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B)\n        or P(AB) denotes the probability of occurrence of both\n        the event &#8216;A&#8217; and &#8216;B&#8217;.<\/p>\n       \n        <p >\n        <p><h4>Question<\/h4> The\n        probability that a contractor will get a contract is\n        &#8216;2\/3&#8217; and the probability that he will get on\n        other contract is 5\/9 . If the probability of getting at\n        least one contract is 4\/5, what is the probability that\n        he will get both the contracts ?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Here P(A) = 2\/3,\n        P(B) = 5\/9<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">b) = 4\/5, (P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B) = ?<\/p>\n        <p >\n        By addition theorem of Probability:<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">B) = P(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        = 4\/5 = 2\/3 + 5\/9\n        &#8211; P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\"\n        height=\"12\">B)<\/p>\n        <p >\n        or 4\/5 = 11\/9 &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        or P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\"\n        width=\"10\" height=\"12\">B) = 11\/9 &#8211; 4\/5\n        = (55-36) \/ 45<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\"\n        width=\"10\" height=\"12\">B) = 19\/45<\/p>\n        <p >\n        <p><h4>Question<\/h4> Two cards are drawn at\n        random. Find the probability that both the cards are of\n        red colour or they are queen.<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Let S =\n        Sample &#8211; space.<\/p>\n        <p >\n       \n        A = The event that the two cards drawn are\n        red.<\/p>\n        <p >\n       \n        B = The event that the two cards drawn are\n        queen.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\"\n        height=\"12\">B = The event that the two cards drawn are\n        queen of red colour.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = <sup>52<\/sup>C<sub>2<\/sub>,\n        n(A) = <sup>26<\/sup>C<sub>2<\/sub>, n(B) = <sup>4<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        n(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\"\n        width=\"10\" height=\"12\">B) = <sup>2<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(A) = n(A) \/ n(S) = <sup>26<\/sup>C<sub>2<\/sub>\n        \/ <sup>52<\/sup>C<sub>2<\/sub> , P(B) = n(B) \/\n        n(S) = <sup>4<\/sup>C<sub>2<\/sub> \/ <sup>52<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\"\n        width=\"10\" height=\"12\">B) = n(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B) \/\n        n(S) = <sup>2<\/sup>C<sub>2<\/sub> \/ <sup>52<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">B) = ?<\/p>\n        <p >\n        We have P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\"\n        width=\"10\" height=\"12\">B) = P(A) + P(B) &#8211; P(A<img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/intersectiongreen.jpg\" width=\"10\" height=\"12\">B)<\/p>\n        <p >\n        = <sup>26<\/sup>C<sub>2<\/sub> \/ <sup>52<\/sup>C<sub>2<\/sub>\n        + <sup>4<\/sup>C<sub>2<\/sub> \/ <sup>52<\/sup>C<sub>2<\/sub>\n        &#8211; <sup>2<\/sup>C<sub>2<\/sub> \/ <sup>52<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        = (<sup>26<\/sup>C<sub>2<\/sub> + <sup>4<\/sup>C<sub>2<\/sub>\n        &#8211; <sup>2<\/sup>C<sub>2<\/sub>) \/ <sup>52<\/sup>C<sub>2<\/sub><\/p>\n        <p >\n        = (13X25+2X3-1) \/ (26X51)<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">B) = 55\/221<\/p>\n        <p >\n        <p><h4>Question<\/h4> A bag contains\n        &#8216;6&#8217; white and &#8216;4&#8217; red balls. Two\n        balls are drawn at random. What is the chance, they will\n        be of the same colour?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Let\n        S = Sample space<\/p>\n        <p >\n       \n        A = the event of drawing &#8216;2&#8217; white balls.<\/p>\n        <p >\n       \n        B = the event of drawing &#8216;2&#8217; red balls.<\/p>\n        <p >\n       \n        A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\">B = The\n        event of drawing 2 white balls or 2 red balls.<\/p>\n        <p >\n       \n        i.e. the event of drawing &#8216;2&#8217; balls of same\n        colour.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> n(S) = <sup>10<\/sup>C<sub>2<\/sub> = 10!\n        \/ (2! x 8!) = 45<\/p>\n        <p >\n        n(A) = <sup>6<\/sup>C<sub>2<\/sub> =\n        6! \/ ((2! x 4!) = (6 x 5 ) \/ 2 = 15<\/p>\n        <p >\n        n(B) = <sup>4<\/sup>C<sub>2<\/sub> =\n        4! \/ (2! x 2!) = (4&#215;3) \/ 2 = 6<\/p>\n        <p >\n        P(A) = n(A) \/ n(S) =\n        15\/45 = 1\/3<\/p>\n        <p >\n        P(B) = n(B) \/ n(S)\n        = 6\/45 = 2\/15<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">B) = P(A) + P(B)<\/p>\n        <p >\n        = 1\/3 + 2\/15\n        = (5+2) \/ 15<\/p>\n        <p >\n        P(A<img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\"\n        height=\"12\">B) = 7\/15<\/p>\n        <p >\n        <p><h4>Question<\/h4> For a\n        post three persons &#8216;A&#8217;, &#8216;B&#8217; and\n        &#8216;C&#8217; appear in the interview. The probability of\n        &#8216;A&#8217; being selected is twice that of\n        &#8216;B&#8217; and the probability of &#8216;B&#8217; being\n        selected is thrice that of &#8216;C&#8217;, what are the\n        individual probability of A, B, C being selected?<\/p>\n        <p >\n        <p><h4>Solution<\/h4> Let &#8216;E<sub>1<\/sub>&#8216;,\n        &#8216;E<sub>2<\/sub>&#8216;, &#8216;E<sub>3<\/sub>&#8216; be\n        the events of selections of A, B, and C respectively.<\/p>\n        <p >\n        Let P(E<sub>3<\/sub>) = x<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>2<\/sub>) = 3. P(E<sub>3<\/sub>)\n        = 3x<\/p>\n        <p >\n        and P(E<sub>1<\/sub>) = 2P(E<sub>2<\/sub>)\n        = 2 x 3x = 6x<\/p>\n        <p >\n        As there are only &#8216;3&#8217;\n        candidates &#8216;A&#8217;, &#8216;B&#8217; and &#8216;C&#8217;\n        we have to select at least one of the candidates A or B\n        or C, surely.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P( E<sub>1<\/sub> <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\"\n        width=\"10\" height=\"12\"> E<sub>2<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\"> E<sub>3<\/sub>)\n        = 1<\/p>\n        <p >\n        and E<sub>1<\/sub>, E<sub>2<\/sub>, E<sub>3<\/sub>\n        are mutually exclusive.<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\">P(E<sub>1<\/sub><img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\"\n        width=\"10\" height=\"12\"> E<sub>2<\/sub> <img loading=\"lazy\" decoding=\"async\"\n        src=\"\/wp-content\/uploads\/2024\/03\/uniongreen.jpg\" width=\"10\" height=\"12\"> E<sub>3<\/sub>)\n        = P(E<sub>1<\/sub>) + P(E<sub>2<\/sub>) + P(E<sub>3<\/sub>)<\/p>\n        <p >\n        1 = 6x + 3x + x<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> 10x &#8211; 1 or x = 1\/10<\/p>\n        <p >\n        <img loading=\"lazy\" decoding=\"async\" src=\"\/wp-content\/uploads\/2024\/03\/arrow.jpg\" width=\"13\"\n        height=\"12\"> P(E<sub>3<\/sub>) = 1\/10, P(E<sub>2<\/sub>)\n        = 3\/10 and P(E<sub>1<\/sub>) = 6\/10\n        = 3\/5<\/p>\n        size=&#8221;4&#8243; face=&#8221;Arial&#8221;><br>\n       \n        <\/p>\n\t\t<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Random Experiment An experiment is said to be a random experiment, if it&#8217;s out-come can&#8217;t be predicted with certainty. Example If a coin is tossed, we can&#8217;t say, whether head or tail will appear. So it is a random experiment. Sample Space The set of all possible out-comes of an experiment is called the sample [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-281","page","type-page","status-publish","hentry"],"blocksy_meta":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Probability: basic concepts - Tutors 4 You<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tutors4you.com\/index.php\/probability-tutorial\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Probability: basic concepts - Tutors 4 You\" \/>\n<meta property=\"og:description\" content=\"Random Experiment An experiment is said to be a random experiment, if it&#8217;s out-come can&#8217;t be predicted with certainty. 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