{"id":42,"date":"2021-01-12T22:19:26","date_gmt":"2021-01-12T22:19:26","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/two-basic-rules-of-probability\/"},"modified":"2021-05-11T20:08:27","modified_gmt":"2021-05-11T20:08:27","slug":"two-basic-rules-of-probability","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/two-basic-rules-of-probability\/","title":{"rendered":"Two Basic Rules of Probability"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.<\/p>\r\n\r\n

\r\n

The Multiplication Rule<\/h3>\r\n

If A<\/em> and B<\/em> are two events defined on a sample space<\/span>, then: P<\/em>(A<\/em> and B<\/em>) = P<\/em>(B<\/em>)P<\/em>(A<\/em>|B<\/em>).<\/p>\r\n

This rule may also be written as: $P(A \\vert B) = \\dfrac{P(A \\text{ and } B)}{P(B)}$<\/p>\r\n

(The probability of A<\/em> given B<\/em> equals the probability of A<\/em> and B<\/em> divided by the probability of B<\/em>.)<\/p>\r\n

If A<\/em> and B<\/em> are independent<\/span>, then P<\/em>(A<\/em>|B<\/em>) = P<\/em>(A<\/em>). Then P<\/em>(A<\/em> and B<\/em>) = P<\/em>(A<\/em>|B<\/em>)P<\/em>(B<\/em>) becomes P<\/em>(A<\/em> and B<\/em>) = P<\/em>(A<\/em>)P<\/em>(B<\/em>).<\/p>\r\n\r\n<\/section>

\r\n

The Addition Rule<\/h3>\r\n

If A<\/em> and B<\/em> are defined on a sample space, then: P<\/em>(A<\/em> or B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>) - P<\/em>(A<\/em> and B<\/em>).<\/p>\r\n

If A<\/em> and B<\/em> are mutually exclusive<\/span>, then P<\/em>(A<\/em> and B<\/em>) = 0. Then P<\/em>(A<\/em> or B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>) - P<\/em>(A<\/em> and B<\/em>) becomes P<\/em>(A<\/em> or B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>).<\/p>\r\n\r\n

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Example <\/span>3.14<\/span><\/h3>\r\n<\/header>
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Klaus is trying to choose where to go on vacation. His two choices are: A<\/em> = New Zealand and B<\/em> = Alaska<\/p>\r\n\r\n

    \r\n \t
  • Klaus can only afford one vacation. The probability that he chooses A<\/em> is P<\/em>(A<\/em>) = 0.6 and the probability that he chooses B<\/em> is P<\/em>(B<\/em>) = 0.35.<\/li>\r\n \t
  • P<\/em>(A<\/em> and B<\/em>) = 0 because Klaus can only afford to take one vacation<\/li>\r\n \t
  • Therefore, the probability that he chooses either New Zealand or Alaska is P<\/em>(A<\/em> or B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n
    \r\n

    Example <\/span>3.15<\/span><\/h3>\r\n<\/header>
    \r\n

    Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A<\/em> = the event Carlos is successful on his first attempt. P<\/em>(A<\/em>) = 0.65. B<\/em> = the event Carlos is successful on his second attempt. P<\/em>(B<\/em>) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN<\/strong> that he made the first goal is 0.90.<\/p>\r\n \r\n

    <\/header>
    \r\n
    \r\n
    \r\n

    a. What is the probability that he makes both goals?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    <\/div>\r\n
    \r\n
    Solution <\/span>3.15<\/span><\/div>\r\n
    \r\n

    a. The problem is asking you to find P<\/em>(A<\/em> and B<\/em>) = P<\/em>(B<\/em> and A<\/em>). Since P<\/em>(B<\/em>|A<\/em>) = 0.90: P<\/em>(B<\/em> and A<\/em>) = P<\/em>(B<\/em>|A<\/em>) P<\/em>(A<\/em>) = (0.90)(0.65) = 0.585<\/p>\r\n

    Carlos makes the first and second goals with probability 0.585.<\/p>\r\n \r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    <\/header>
    \r\n
    \r\n
    \r\n

    b. What is the probability that Carlos makes either the first goal or the second goal?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    <\/div>\r\n
    \r\n
    Solution <\/span>3.15<\/span><\/div>\r\n
    \r\n

    b. The problem is asking you to find P<\/em>(A<\/em> or B<\/em>).<\/p>\r\n

    P<\/em>(A<\/em> or B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>) - P<\/em>(A<\/em> and B<\/em>) = 0.65 + 0.65 - 0.585 = 0.715<\/p>\r\n

    Carlos makes either the first goal or the second goal with probability 0.715.<\/p>\r\n \r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    <\/header>
    \r\n
    \r\n
    \r\n

    c. Are A<\/em> and B<\/em> independent?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    <\/div>\r\n
    \r\n
    Solution <\/span>3.15<\/span><\/div>\r\n
    \r\n

    c. No, they are not, because P<\/em>(B<\/em> and A<\/em>) = 0.585.<\/p>\r\n

    P<\/em>(B<\/em>)P<\/em>(A<\/em>) = (0.65)(0.65) = 0.423<\/p>\r\n

    0.423 \u2260 0.585 = P<\/em>(B<\/em> and A<\/em>)<\/p>\r\n

    So, P<\/em>(B<\/em> and A<\/em>) is not<\/strong> equal to P<\/em>(B<\/em>)P<\/em>(A<\/em>).<\/p>\r\n \r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    <\/header>
    \r\n
    \r\n
    \r\n

    d. Are A<\/em> and B<\/em> mutually exclusive?<\/p>\r\n\r\n

    Solution <\/span>3.15<\/span><\/h4>\r\n
    \r\n
      \r\n \t
    1. The problem is asking you to find P<\/em>(A<\/em> AND B<\/em>) = P<\/em>(B<\/em> AND A<\/em>). Since P<\/em>(B<\/em>|A<\/em>) = 0.90: P<\/em>(B<\/em> AND A<\/em>) = P<\/em>(B<\/em>|A<\/em>) P<\/em>(A<\/em>)\u00a0=\u00a0(0.90)(0.65) = 0.585\r\nCarlos makes the first and second goals with probability 0.585.\r\n<\/span><\/li>\r\n \t
    2. \r\n

      The problem is asking you to find P<\/em>(A<\/em> OR B<\/em>).<\/p>\r\n

      P<\/em>(A<\/em> OR B<\/em>) = P<\/em>(A<\/em>) + P<\/em>(B<\/em>) - P<\/em>(A<\/em> AND B<\/em>) = 0.65 + 0.65 - 0.585 = 0.715<\/p>\r\n

      Carlos makes either the first goal or the second goal with probability 0.715.\r\n<\/span><\/p>\r\n<\/li>\r\n \t

    3. \r\n

      No, they are not, because P<\/em>(B<\/em> AND A<\/em>) = 0.585.<\/p>\r\n

      P<\/em>(B<\/em>)P<\/em>(A<\/em>) = (0.65)(0.65) = 0.423<\/p>\r\n

      0.423 \u2260 0.585 = P<\/em>(B<\/em> AND A<\/em>)<\/p>\r\n

      So, P<\/em>(B<\/em> AND A<\/em>) is not<\/strong> equal to P<\/em>(B<\/em>)P<\/em>(A<\/em>).<\/p>\r\n<\/li>\r\n \t

    4. \r\n

      No, they are not because P<\/em>(A<\/em> and B<\/em>) = 0.585.<\/p>\r\n

      To be mutually exclusive, P<\/em>(A<\/em> AND B<\/em>) must equal zero.<\/p>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      <\/div>\r\n
      \r\n
      Solution <\/span>3.15<\/span><\/div>\r\n
      \r\n

      d. No, they are not because P<\/em>(A<\/em> and B<\/em>) = 0.585.<\/p>\r\n

      To be mutually exclusive, P<\/em>(A<\/em> and B<\/em>) must equal zero.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

      \r\n

      Try It <\/span>3.15<\/span><\/h3>\r\n<\/header>
      \r\n
      <\/header>
      \r\n
      \r\n
      \r\n

      Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C<\/em> = the event that Helen makes the first shot. P<\/em>(C<\/em>) = 0.75. D<\/em> = the event Helen makes the second shot. P<\/em>(D<\/em>) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

      \r\n

      Example <\/span>3.16<\/span><\/h3>\r\n<\/header>
      \r\n

      A community swim team has 150<\/strong> members. Seventy-five<\/strong> of the members are advanced swimmers. Forty-seven<\/strong> of the members are intermediate swimmers. The remainder are novice swimmers. Forty<\/strong> of the advanced swimmers practice four times a week. Thirty<\/strong> of the intermediate swimmers practice four times a week. Ten<\/strong> of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.<\/p>\r\n \r\n

      <\/header>
      \r\n
      \r\n
      \r\n

      a. What is the probability that the member is a novice swimmer?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      <\/div>\r\n
      \r\n
      Solution <\/span>3.16<\/span><\/div>\r\n
      \r\n

      a. 28<\/span><\/span>150<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n2815028150