{"id":47,"date":"2021-06-22T15:30:13","date_gmt":"2021-06-22T15:30:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-basic-rules-of-probability\/"},"modified":"2023-12-05T09:02:35","modified_gmt":"2023-12-05T09:02:35","slug":"two-basic-rules-of-probability","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-basic-rules-of-probability\/","title":{"raw":"Rules for Finding Probabilities","rendered":"Rules for Finding Probabilities"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Calculate probabilities based on the addition rule<\/li>\r\n \t
  • Calculate probabilities based on the multiplication rule<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nWhen 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.\r\n

    The Multiplication Rule<\/h2>\r\nIf\u00a0[latex]A[\/latex] and [latex]B[\/latex] are two events defined on a sample space<\/strong>, then: [latex]P(A \\text{ AND } B) = P(B)P(A|B)[\/latex].\r\n\r\nThis rule may also be written as\u00a0[latex]\\displaystyle{P}{({A}{\\mid}{B})}=\\frac{{{P}{({A}\\text{ AND } {B})}}}{{{P}{({B})}}}[\/latex]\r\n\r\n(The probability of\u00a0[latex]A[\/latex] given [latex]B[\/latex] equals the probability of [latex]A[\/latex] and [latex]B[\/latex] divided by the probability of [latex]B[\/latex].)\r\n\r\nIf\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent<\/strong>, then [latex]P(A|B) = P(A)[\/latex]. Then [latex]P(A \\text{ AND } B) = P(A|B)P(B)[\/latex] becomes [latex]P(A \\text{ AND } B) = P(A)P(B)[\/latex].\r\n

    The Addition Rule<\/h2>\r\nIf\u00a0[latex]A[\/latex] and [latex]B[\/latex] are defined on a sample space, then: [latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B)[\/latex].\r\n\r\nIf\u00a0[latex]A[\/latex] and [latex]B[\/latex] are mutually exclusive<\/strong>, then [latex]P(A \\text{ AND } B) = 0[\/latex]. Then [latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B)[\/latex] becomes [latex]P(A \\text{ OR } B) = P(A) + P(B)[\/latex].\r\n

    <\/h3>\r\n
    \r\n

    Example<\/h3>\r\nKlaus is trying to choose where to go on vacation. His two choices are: [latex]A[\/latex] = New Zealand and [latex]B[\/latex] = Alaska.\r\n
      \r\n \t
    • Klaus can only afford one vacation. The probability that he chooses [latex]A[\/latex] is [latex]P(A) = 0.6[\/latex] and the probability that he chooses [latex]B[\/latex] is [latex]P(B) = 0.35[\/latex].<\/li>\r\n \t
    • [latex]P(A \\text{ AND } B) = 0[\/latex] 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 [latex]P(A \\text{ OR } B) = P(A) + P(B) = 0.6 + 0.35 = 0.95[\/latex]. Note that the probability that he does not choose to go anywhere on vacation must be [latex]0.05[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n
      \r\n

      Recall: Add or Subtract Decimals<\/h3>\r\n
        \r\n \t
      1. Write the numbers vertically so the decimal points line up.<\/li>\r\n \t
      2. Use zeros as place holders, as needed.<\/li>\r\n \t
      3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Example<\/h3>\r\nCarlos plays college soccer. He makes a goal [latex]65[\/latex]% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. [latex]A =[\/latex] the event Carlos is successful on his first attempt. [latex]P(A) = 0.65. B = [\/latex] the event Carlos is successful on his second attempt. [latex]P(B) = 0.65[\/latex]. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is [latex]0.90[\/latex].\r\n
          \r\n \t
        1. What is the probability that he makes both goals?<\/li>\r\n \t
        2. What is the probability that Carlos makes either the first goal or the second goal?<\/li>\r\n \t
        3. Are [latex]A[\/latex] and [latex]B[\/latex] independent?<\/li>\r\n \t
        4. Are [latex]A[\/latex] and [latex]B[\/latex] mutually exclusive?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124075\"]\r\n\r\nSolution:\r\n
            \r\n \t
          1. The problem is asking you to find [latex]P(A \\text{ AND } B) = P(B \\text{ AND } A)[\/latex]. Since [latex]P(B|A) = 0.90: P(B \\text{ AND } A) = P(B|A) P(A) = (0.90)(0.65) = 0.585[\/latex]. Carlos makes the first and second goals with probability [latex]0.585[\/latex].<\/li>\r\n \t
          2. The problem is asking you to find [latex]P(A \\text{ OR } B)[\/latex].\r\n[latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B) = 0.65 + 0.65 - 0.585 = 0.715[\/latex]. Carlos makes either the first goal or the second goal with probability [latex]0.715[\/latex].<\/li>\r\n \t
          3. No, they are not, because [latex]P(B \\text{ AND } A) = 0.585[\/latex].[latex]P(B)P(A) = (0.65)(0.65) = 0.423 \\, 0.423 \\neq 0.585 = P(B \\text{ AND } A){.}[\/latex] So, [latex]P(B \\text{ AND } A)[\/latex] is NOT<\/strong> equal to [latex]P(B)P(A)[\/latex].<\/li>\r\n \t
          4. No, they are not because [latex]P(A \\text{ AND } B) = 0.585[\/latex]. To be mutually exclusive, [latex]P(A \\text{ AND } B)[\/latex] must equal zero.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\n
            \r\n\r\nWatch this video for another example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome.\r\n\r\nhttps:\/\/youtu.be\/z-1VvourLsA\r\n

            <\/h3>\r\n
            \r\n

            Try it<\/h3>\r\nHelen plays basketball. For free throws, she makes the shot [latex]75[\/latex]% of the time. Helen must now attempt two free throws.\u00a0[latex]C[\/latex] = the event that Helen makes the first shot. [latex]P(C) = 0.75[\/latex]. [latex]D[\/latex] = the event Helen makes the second shot. [latex]P(D) = 0.75[\/latex]. The probability that Helen makes the second free throw given that she made the first is [latex]0.85[\/latex]. What is the probability that Helen makes both free throws?\r\n\r\n[reveal-answer q=\"124076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124076\"]\r\n\r\n[latex]P(D|C) = 0.85[\/latex]\r\n\r\n[latex]P(C \\ \\text{AND} \\ D) = P(D \\ \\text{AND} \\ C)[\/latex]\r\n\r\n[latex]P(D \\ \\text{AND} \\ C) = P(D|C)P(C) = (0.85)(0.75) = 0.6375[\/latex]\r\n\r\nHelen makes the first and second free throws with probability [latex]0.6375[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            <\/h3>\r\n
            \r\n

            Example<\/h3>\r\nA community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.\r\n
              \r\n \t
            1. What is the probability that the member is a novice swimmer?<\/li>\r\n \t
            2. What is the probability that the member practices four times a week?<\/li>\r\n \t
            3. What is the probability that the member is an advanced swimmer and practices four times a week?<\/li>\r\n \t
            4. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?<\/li>\r\n \t
            5. Are being a novice swimmer and practicing four times a week independent events? Why or why not?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124077\"]\r\nSolution:\r\n
                \r\n \t
              1. [latex]\\displaystyle\\frac{{28}}{{150}}[\/latex]<\/li>\r\n \t
              2. [latex]\\displaystyle\\frac{{80}}{{150}}[\/latex]<\/li>\r\n \t
              3. [latex]\\displaystyle\\frac{{40}}{{150}}[\/latex]<\/li>\r\n \t
              4. [latex]P[\/latex](advanced AND intermediate) = [latex]0[\/latex], so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.<\/li>\r\n \t
              5. No, these are not independent events. [latex]P[\/latex](novice AND practices four times per week) = [latex]0.0667\\,\\,\\,\\,P[\/latex](novice)[latex]P[\/latex](practices four times per week) [latex]= 0.0996\\,\\,\\,\\,0.0667 \\neq 0.0996[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                <\/h3>\r\n
                \r\n

                try it<\/h3>\r\nA school has [latex]200[\/latex] seniors, of whom [latex]140[\/latex] will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?\r\n[reveal-answer q=\"124078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124078\"]\r\n[latex]\\displaystyle{P}=\\frac{{{200}-{140}-{40}}}{{200}}=\\frac{{20}}{{200}}={0.1}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                <\/h3>\r\n
                \r\n

                Example<\/h3>\r\nFelicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class, given that she enrolls in speech class, is 0.25.\r\n\r\nLet\u00a0[latex]M[\/latex] = math class, [latex]S[\/latex] = speech class, [latex]M|S[\/latex] = math given speech.\r\n
                  \r\n \t
                1. What is the probability that Felicity enrolls in math and speech? Find [latex]P(M \\text{ AND } S) = P(M|S)P(S)[\/latex].<\/li>\r\n \t
                2. What is the probability that Felicity enrolls in math or speech classes? Find [latex]P(M \\text{ OR } S) = P(M) + P(S) - P(M \\text{ AND } S)[\/latex].<\/li>\r\n \t
                3. Are [latex]M[\/latex] and [latex]S[\/latex] independent? Is [latex]P(M|S) = P(M)[\/latex]?<\/li>\r\n \t
                4. Are [latex]M[\/latex] and [latex]S[\/latex] mutually exclusive? Is [latex]P(M \\text{ AND } S) = 0[\/latex]?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124079\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124079\"]\r\nSolution:\r\n
                    \r\n \t
                  1. [latex]0.1625[\/latex]<\/li>\r\n \t
                  2. [latex]0.6875[\/latex]<\/li>\r\n \t
                  3. No<\/li>\r\n \t
                  4. No<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n
                    \r\n

                    try it<\/h3>\r\nA student goes to the library. Let events\u00a0[latex]B[\/latex] = the student checks out a book and [latex]D[\/latex] = the student check out a DVD. Suppose that [latex]P(B) = 0.40[\/latex], [latex]P(D) = 0.30[\/latex] and [latex]P(D|B) = 0.5[\/latex].\r\n
                      \r\n \t
                    1. Find [latex]P(B \\text{ AND } D)[\/latex].<\/li>\r\n \t
                    2. Find [latex]P(B \\text{ OR } D)[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124080\"]\r\n
                        \r\n \t
                      1. [latex]P(B \\text{ AND } D) = P(D|B)P(B) = (0.5)(0.4) = 0.20.[\/latex]<\/li>\r\n \t
                      2. [latex]P(B \\text{ OR } D) = P(B) + P(D) \u2212 P(B \\text{ AND } D) = 0.40 + 0.30 \u2212 0.20 = 0.50[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                        <\/h3>\r\n
                        \r\n

                        Example<\/h3>\r\nStudies show that about one woman in seven (approximately [latex]14.3[\/latex]%) who live to be [latex]90[\/latex] will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative [latex]2[\/latex]% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about [latex]85[\/latex]% of the time. Let [latex]B[\/latex] = woman develops breast cancer and let [latex]N[\/latex] = tests negative. Suppose one woman is selected at random.\r\n
                          \r\n \t
                        1. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?<\/li>\r\n \t
                        2. Given that the woman has breast cancer, what is the probability that she tests negative?<\/li>\r\n \t
                        3. What is the probability that the woman has breast cancer AND tests negative?<\/li>\r\n \t
                        4. What is the probability that the woman has breast cancer or tests negative?<\/li>\r\n \t
                        5. Are having breast cancer and testing negative independent events?<\/li>\r\n \t
                        6. Are having breast cancer and testing negative mutually exclusive?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124081\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124081\"]\r\nSolution:\r\n
                            \r\n \t
                          1. [latex]P(B) = 0.143; P(N) = 0.85[\/latex]<\/li>\r\n \t
                          2. [latex]P(N|B) = 0.02[\/latex]<\/li>\r\n \t
                          3. [latex]P(B \\text{ AND } N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029[\/latex]<\/li>\r\n \t
                          4. [latex]P(B \\text{ OR } N) = P(B) + P(N) - P(B \\text{ AND } N) = 0.143 + 0.85 - 0.0029 = 0.9901[\/latex]<\/li>\r\n \t
                          5. No.\u00a0[latex]P(N) = 0.85; P(N|B) = 0.02[\/latex]. So, [latex]P(N|B)[\/latex] does not equal [latex]P(N)[\/latex].<\/li>\r\n \t
                          6. No.\u00a0[latex]P(B \\text{ AND } N) = 0.0029[\/latex]. For [latex]B[\/latex] and [latex]N[\/latex] to be mutually exclusive, [latex]P(B \\text{ AND } N)[\/latex] must be zero.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n
                            \r\n

                            Try it<\/h3>\r\nA school has [latex]200[\/latex] seniors, of whom [latex]140[\/latex] will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?\r\n\r\n[reveal-answer q=\"124082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124082\"]\r\n\r\nLet\u00a0[latex]A[\/latex] = student is a senior going to college.\r\n\r\nLet\u00a0[latex]B[\/latex] = student plays sports.\r\n\r\n[latex]\\displaystyle{P}{({B})}=\\frac{{140}}{{200}}[\/latex]\r\n\r\n[latex]\\displaystyle{P}{({B}|{A})}=\\frac{{50}}{{140}}[\/latex]\r\n\r\n[latex]P(A \\text{ AND } B) = P(B|A)P(A)[\/latex]\r\n\r\n[latex]\\displaystyle{P}{({A} \\text{ AND } {B})}={(\\frac{{140}}{{200}})}{(\\frac{{50}}{{140}})}=\\frac{{1}}{{4}}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                            <\/h3>\r\n
                            \r\n

                            Example<\/h3>\r\nRefer to the information in the earlier example on women who develop breast cancer. [latex]P[\/latex] = tests positive.\r\n
                              \r\n \t
                            1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find [latex]P(P|B) = 1 \u2013 P(N|B)[\/latex].<\/li>\r\n \t
                            2. What is the probability that a woman develops breast cancer and tests positive? Find [latex]P(B \\text{ AND } P) = P(P|B)P(B)[\/latex].<\/li>\r\n \t
                            3. What is the probability that a woman does not develop breast cancer? Find [latex]P(B') = 1 \u2013 P(B)[\/latex].<\/li>\r\n \t
                            4. What is the probability that a woman tests positive for breast cancer? Find [latex]P(P) = 1 \u2013 P(N)[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124083\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124083\"]\r\nSolution:\r\n
                                \r\n \t
                              1. [latex]0.98[\/latex]<\/li>\r\n \t
                              2. [latex]0.1401[\/latex]<\/li>\r\n \t
                              3. [latex]0.857[\/latex]<\/li>\r\n \t
                              4. [latex]0.15[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                                <\/h4>\r\n
                                \r\n

                                try it<\/h3>\r\nA student goes to the library. Let events\u00a0[latex]B[\/latex] = the student checks out a book and [latex]D[\/latex] = the student checks out a DVD. Suppose that [latex]P(B) = 0.40[\/latex], [latex]P(D) = 0.30[\/latex] and [latex]P(D|B) = 0.5[\/latex].\r\n
                                  \r\n \t
                                1. Find [latex]P(B')[\/latex].<\/li>\r\n \t
                                2. Find [latex]P(D \\text{ AND } B)[\/latex].<\/li>\r\n \t
                                3. Find [latex]P(B|D)[\/latex].<\/li>\r\n \t
                                4. Find [latex]P(D \\text{ AND } B')[\/latex].<\/li>\r\n \t
                                5. Find [latex]P(D|B')[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124084\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124084\"]\r\n
                                    \r\n \t
                                  1. [latex]P(B') = 0.60[\/latex]<\/li>\r\n \t
                                  2. [latex]P(D \\text{ AND } B) = P(D|B)P(B) = 0.20[\/latex]<\/li>\r\n \t
                                  3. [latex]P(B|D)=\\displaystyle\\frac{{{P}{({B}\\text{ AND } {D})}}}{{{P}{({D})}}}=\\frac{{{0.20}}}{{{0.30}}}={0.66}[\/latex]<\/li>\r\n \t
                                  4. [latex]P(D \\text{ AND } B') = P(D) - P(D \\text{ AND } B) = 0.30 - 0.20 = 0.10[\/latex]<\/li>\r\n \t
                                  5. [latex]P(D|B') = P(D \\text{ AND } B')P(B') = (P(D) - P(D \\text{ AND } B))(0.60) = (0.10)(0.60) = 0.06[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"
                                    \n

                                    Learning Outcomes<\/h3>\n
                                    \n
                                      \n
                                    • Calculate probabilities based on the addition rule<\/li>\n
                                    • Calculate probabilities based on the multiplication rule<\/li>\n<\/ul>\n<\/section>\n<\/div>\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>\n

                                      The Multiplication Rule<\/h2>\n

                                      If\u00a0[latex]A[\/latex] and [latex]B[\/latex] are two events defined on a sample space<\/strong>, then: [latex]P(A \\text{ AND } B) = P(B)P(A|B)[\/latex].<\/p>\n

                                      This rule may also be written as\u00a0[latex]\\displaystyle{P}{({A}{\\mid}{B})}=\\frac{{{P}{({A}\\text{ AND } {B})}}}{{{P}{({B})}}}[\/latex]<\/p>\n

                                      (The probability of\u00a0[latex]A[\/latex] given [latex]B[\/latex] equals the probability of [latex]A[\/latex] and [latex]B[\/latex] divided by the probability of [latex]B[\/latex].)<\/p>\n

                                      If\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent<\/strong>, then [latex]P(A|B) = P(A)[\/latex]. Then [latex]P(A \\text{ AND } B) = P(A|B)P(B)[\/latex] becomes [latex]P(A \\text{ AND } B) = P(A)P(B)[\/latex].<\/p>\n

                                      The Addition Rule<\/h2>\n

                                      If\u00a0[latex]A[\/latex] and [latex]B[\/latex] are defined on a sample space, then: [latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B)[\/latex].<\/p>\n

                                      If\u00a0[latex]A[\/latex] and [latex]B[\/latex] are mutually exclusive<\/strong>, then [latex]P(A \\text{ AND } B) = 0[\/latex]. Then [latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B)[\/latex] becomes [latex]P(A \\text{ OR } B) = P(A) + P(B)[\/latex].<\/p>\n

                                      <\/h3>\n
                                      \n

                                      Example<\/h3>\n

                                      Klaus is trying to choose where to go on vacation. His two choices are: [latex]A[\/latex] = New Zealand and [latex]B[\/latex] = Alaska.<\/p>\n

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

                                        Recall: Add or Subtract Decimals<\/h3>\n
                                          \n
                                        1. Write the numbers vertically so the decimal points line up.<\/li>\n
                                        2. Use zeros as place holders, as needed.<\/li>\n
                                        3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.<\/li>\n<\/ol>\n<\/div>\n
                                          \n

                                          Example<\/h3>\n

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

                                            \n
                                          1. What is the probability that he makes both goals?<\/li>\n
                                          2. What is the probability that Carlos makes either the first goal or the second goal?<\/li>\n
                                          3. Are [latex]A[\/latex] and [latex]B[\/latex] independent?<\/li>\n
                                          4. Are [latex]A[\/latex] and [latex]B[\/latex] mutually exclusive?<\/li>\n<\/ol>\n
                                            Show Solution<\/span><\/p>\n
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                                            Solution:<\/p>\n

                                              \n
                                            1. The problem is asking you to find [latex]P(A \\text{ AND } B) = P(B \\text{ AND } A)[\/latex]. Since [latex]P(B|A) = 0.90: P(B \\text{ AND } A) = P(B|A) P(A) = (0.90)(0.65) = 0.585[\/latex]. Carlos makes the first and second goals with probability [latex]0.585[\/latex].<\/li>\n
                                            2. The problem is asking you to find [latex]P(A \\text{ OR } B)[\/latex].
                                              \n[latex]P(A \\text{ OR } B) = P(A) + P(B) - P(A \\text{ AND } B) = 0.65 + 0.65 - 0.585 = 0.715[\/latex]. Carlos makes either the first goal or the second goal with probability [latex]0.715[\/latex].<\/li>\n
                                            3. No, they are not, because [latex]P(B \\text{ AND } A) = 0.585[\/latex].[latex]P(B)P(A) = (0.65)(0.65) = 0.423 \\, 0.423 \\neq 0.585 = P(B \\text{ AND } A){.}[\/latex] So, [latex]P(B \\text{ AND } A)[\/latex] is NOT<\/strong> equal to [latex]P(B)P(A)[\/latex].<\/li>\n
                                            4. No, they are not because [latex]P(A \\text{ AND } B) = 0.585[\/latex]. To be mutually exclusive, [latex]P(A \\text{ AND } B)[\/latex] must equal zero.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n

                                               <\/p>\n

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                                              Watch this video for another example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome.<\/p>\n