{"id":51,"date":"2021-06-22T15:30:15","date_gmt":"2021-06-22T15:30:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-13\/"},"modified":"2023-12-05T09:06:33","modified_gmt":"2023-12-05T09:06:33","slug":"answers-to-selected-exercises-13","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-13\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"1.\r\n\r\n
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    \r\n \t
  1. P<\/em>(L\u2032<\/em>) = P<\/em>(S<\/em>)<\/li>\r\n \t
  2. P<\/em>(M<\/em> OR S<\/em>)<\/li>\r\n \t
  3. P<\/em>(F<\/em> AND L<\/em>)<\/li>\r\n \t
  4. P<\/em>(M<\/em>|L<\/em>)<\/li>\r\n \t
  5. P<\/em>(L<\/em>|M<\/em>)<\/li>\r\n \t
  6. P<\/em>(S<\/em>|F<\/em>)<\/li>\r\n \t
  7. P<\/em>(F<\/em>|L<\/em>)<\/li>\r\n \t
  8. P<\/em>(F<\/em> OR L<\/em>)<\/li>\r\n \t
  9. P<\/em>(M<\/em> AND S<\/em>)<\/li>\r\n \t
  10. P<\/em>(F<\/em>)<\/li>\r\n<\/ol>\r\n3.\u00a0P<\/em>(N<\/em>) = [latex]\\frac{{15}}{{42}}=\\frac{{5}}{{14}}=0.36[\/latex]\r\n\r\n5.\u00a0P<\/em>(C<\/em>) = [latex]\\frac{{5}}{{42}}=0.12[\/latex]\r\n\r\n7.\u00a0P<\/em>(G<\/em>) = [latex]\\frac{{20}}{{150}}=\\frac{{2}}{{15}}=0.13[\/latex]\r\n\r\n9.\u00a0P<\/em>(R<\/em>) = [latex]\\frac{{22}}{{150}}=\\frac{{11}}{{75}}=0.15[\/latex]\r\n\r\n11.\u00a0P<\/em>(O<\/em>) = [latex]\\frac{{{22}-{38}-{20}-{28}-{26}}}{{150}}=\\frac{{16}}{{150}}=\\frac{{8}}{{75}}=0.11[\/latex]\r\n\r\n13.\u00a0P<\/em>(E<\/em>) = [latex]\\frac{{47}}{{194}}=0.24[\/latex]\r\n\r\n15.\u00a0P<\/em>(N<\/em>) = [latex]\\frac{{23}}{{194}}=0.12[\/latex]\r\n\r\n17.\u00a0P<\/em>(S<\/em>)=[latex]\\frac{{12}}{{194}}={{6}}{{97}}=0.06[\/latex]\r\n\r\n19. [latex]\\frac{{13}}{{52}}={{1}}{{4}}=0.25[\/latex]\r\n\r\n21. [latex]\\frac{{3}}{{6}}={{1}}{{2}}=0.5[\/latex]\r\n\r\n23. P(R) = [latex]\\frac{{4}}{{8}}=0.5[\/latex]\r\n\r\n25. P(O or H)\r\n\r\n27. P(H | I)\r\n\r\n29. P(N | O)\r\n\r\n31. P(I or N)\r\n\r\n33. P(I)\r\n\r\n35.\u00a0The likelihood that an event will occur given that another event has already occurred.\r\n\r\n37. 1\r\n\r\n39.\u00a0the probability of landing on an even number or a multiple of three\r\n\r\n<\/section>\r\n

    Independent and Mutually Exclusive Events \u2014 Practice<\/h2>\r\n41. P(J) = 0.3\r\n43. P(Q AND R) = P(Q)P(R)0.1 = (0.4)P(R)P(R) = 0.25\r\n

    Two Basic Rules of Probability \u2014 Practice<\/h2>\r\n45. 0.376\r\n47. C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.\r\n49. L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.\r\n51. 0.6492\r\n53. No, because P(L AND C) does not equal 0.\r\n

    \u00a0Contingency Tables \u2014 Practice<\/h2>\r\n55. P(musician is a male AND had private instruction) = [latex]\\frac{15}{130} = [latex]\\frac{3}{26} = 0.12[\/latex]\r\n57. The events are not mutually exclusive. It is possible to be a female musician who learned music in school.\r\n
    \r\n

    Tree and Venn Diagrams \u2014 Practice<\/h2>\r\n<\/div>\r\n58.\"This\r\n\r\n60. [latex]\\frac{35,065}{100,450}[\/latex]\r\n\r\n62. To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is [latex]\\frac{4,715}{100,450}[\/latex].\r\n\r\n64. To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is [latex]\\frac{4715}{15,273}[\/latex].\r\n
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    \r\n
    \r\n

    Terminology \u2014 Homework<\/h2>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n67.\r\n
      \r\n \t
    1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%<\/li>\r\n \t
    2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.<\/li>\r\n<\/ol>\r\n
      \r\n
      \r\n

      Independent and Mutually Exclusive Events \u2014 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n
      <\/section>69. 0\r\n71. 0.3571\r\n73. 0.2142\r\n75. Physician (83.7)\r\n77. 83.7 \u2212 79.6 = 4.1\r\n79. P(Occupation < 81.3) = 0.5\r\n
      \r\n
      \r\n

      Two Basic Rules of Probability \u2014 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n81.\r\n\r\n
      \r\n
        \r\n \t
      1. The Forum Research surveyed 1,046 Torontonians.<\/li>\r\n \t
      2. 58%<\/li>\r\n \t
      3. 42% of 1,046 = 439 (rounding to the nearest integer)<\/li>\r\n \t
      4. 0.57<\/li>\r\n \t
      5. 0.60.<\/li>\r\n<\/ol>\r\n83.\r\n
          \r\n \t
        1. P<\/em>(Betting on two line that touch each other on the table) = [latex]\\frac{{6}}{{38}}[\/latex]<\/li>\r\n \t
        2. P<\/em>(Betting on three numbers in a line) = [latex]\\frac{{3}}{{38}}[\/latex]<\/li>\r\n \t
        3. P<\/em>(Bettting on one number) = [latex]\\frac{{1}}{{38}}[\/latex]<\/li>\r\n \t
        4. P<\/em>(Betting on four number that touch each other to form a square) = [latex]\\frac{{4}}{{38}}[\/latex]<\/li>\r\n \t
        5. P<\/em>(Betting on two number that touch each other on the table ) = [latex]\\frac{{2}}{{38}}[\/latex]<\/li>\r\n \t
        6. P<\/em>(Betting on 0-00-1-2-3) = [latex]\\frac{{5}}{{38}}[\/latex]<\/li>\r\n \t
        7. P<\/em>(Betting on 0-1-2; or 0-00-2; or 00-2-3) = [latex]\\frac{{3}}{{38}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>85.\r\n\r\n
          \r\n
            \r\n \t
          1. {G<\/em>1, G<\/em>2, G<\/em>3, G<\/em>4, G<\/em>5, Y<\/em>1, Y<\/em>2, Y<\/em>3}<\/li>\r\n \t
          2. [latex]\\frac{{5}}{{8}}[\/latex]<\/li>\r\n \t
          3. [latex]\\frac{{2}}{{3}}[\/latex]<\/li>\r\n \t
          4. [latex]\\frac{{2}}{{8}}[\/latex]<\/li>\r\n \t
          5. [latex]\\frac{{6}}{{8}}[\/latex]<\/li>\r\n \t
          6. No, because P<\/em>(G<\/em> AND E<\/em>) does not equal 0.<\/li>\r\n<\/ol>\r\n87.\r\n
            \r\n
            NOTE<\/div>\r\n<\/header>
            \r\n

            The coin toss is independent of the card picked first.<\/p>\r\n\r\n<\/section><\/div>\r\n

              \r\n \t
            1. {(G<\/em>,H<\/em>) (G<\/em>,T<\/em>) (B<\/em>,H<\/em>) (B<\/em>,T<\/em>) (R<\/em>,H<\/em>) (R<\/em>,T<\/em>)}<\/li>\r\n \t
            2. P<\/em>(A<\/em>) = P<\/em>(blue)P<\/em>(head) = ([latex]\\frac{{3}}{{10}}[\/latex])([latex]\\frac{{1}}{{2}}[\/latex])=[latex]\\frac{{3}}{{20}}[\/latex]<\/li>\r\n \t
            3. Yes, A<\/em> and B<\/em> are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P<\/em>(A<\/em> AND B<\/em>) = 0<\/li>\r\n \t
            4. No, A<\/em> and C<\/em> are not mutually exclusive because they can occur at the same time. In fact, C<\/em> includes all of the outcomes of A<\/em>; if the card chosen is blue it is also (red or blue). P<\/em>(A<\/em> AND C<\/em>) = P<\/em>(A<\/em>) = 3<\/span>20<\/span><\/span><\/span><\/span><\/span><\/span><\/li>\r\n<\/ol>\r\n89.\r\n\r\n
              \r\n
                \r\n \t
              1. S<\/em> = {(HHH<\/em>), (HHT<\/em>), (HTH<\/em>), (HTT<\/em>), (THH<\/em>), (THT<\/em>), (TTH<\/em>), (TTT<\/em>)}<\/li>\r\n \t
              2. [latex]\\frac{{4}}{{8}}[\/latex]<\/li>\r\n \t
              3. Yes, because if A<\/em> has occurred, it is impossible to obtain two tails. In other words, P<\/em>(A<\/em> AND B<\/em>) = 0.<\/li>\r\n<\/ol>\r\n91.\r\n\r\n
                \r\n
                  \r\n \t
                1. If Y<\/em> and Z<\/em> are independent, then P<\/em>(Y<\/em> AND Z<\/em>) = P<\/em>(Y<\/em>)P<\/em>(Z<\/em>), so P<\/em>(Y<\/em> OR Z<\/em>) = P<\/em>(Y<\/em>) + P<\/em>(Z<\/em>) - P<\/em>(Y<\/em>)P<\/em>(Z<\/em>).<\/li>\r\n \t
                2. 0.5<\/li>\r\n<\/ol>\r\n93.\r\n\r\n
                  \r\n

                  1. iii<\/span>\u00a02.\u00a0i<\/span>\u00a03.\u00a0iv<\/span>\u00a04.\u00a0ii<\/span><\/span><\/p>\r\n\r\n<\/section>95.\r\n\r\n

                  \r\n
                    \r\n \t
                  1. P<\/em>(R<\/em>) = 0.44<\/li>\r\n \t
                  2. P<\/em>(R<\/em>|E<\/em>) = 0.56<\/li>\r\n \t
                  3. P<\/em>(R<\/em>|O<\/em>) = 0.31<\/li>\r\n \t
                  4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P<\/em>(R<\/em>|E<\/em>) \u2260 P<\/em>(R<\/em>).<\/li>\r\n \t
                  5. No, this study definitely does not support that notion; in fact<\/u><\/em>, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P<\/em>(R<\/em>|E<\/em>) > P<\/em>(R<\/em>).<\/li>\r\n<\/ol>\r\n<\/section>
                    \u00a097.
                    \r\n
                      \r\n \t
                    1. \r\n

                      P<\/em>(type O OR Rh-) = P<\/em>(type O) + P<\/em>(Rh-) - P<\/em>(type O AND Rh-)<\/p>\r\n

                      0.52 = 0.43 + 0.15 - P<\/em>(type O AND Rh-); solve to find P<\/em>(type O AND Rh-) = 0.06<\/p>\r\n

                      6% of people have type O, Rh- blood<\/p>\r\n<\/li>\r\n \t

                    2. \r\n

                      P<\/em>(NOT(type O AND Rh-)) = 1 - P<\/em>(type O AND Rh-) = 1 - 0.06 = 0.94<\/p>\r\n

                      94% of people do not have type O, Rh- blood<\/p>\r\n<\/li>\r\n<\/ol>\r\n99.\r\n\r\n

                      \r\n
                        \r\n \t
                      1. Let C<\/em> = be the event that the cookie contains chocolate. Let N<\/em> = the event that the cookie contains nuts.<\/li>\r\n \t
                      2. P<\/em>(C<\/em> OR N<\/em>) = P<\/em>(C<\/em>) + P<\/em>(N<\/em>) - P<\/em>(C<\/em> AND N<\/em>) = 0.36 + 0.12 - 0.08 = 0.40<\/li>\r\n \t
                      3. P<\/em>(NEITHER chocolate NOR nuts) = 1 - P<\/em>(C<\/em> OR N<\/em>) = 1 - 0.40 = 0.60<\/li>\r\n<\/ol>\r\n<\/section><\/section><\/section><\/section><\/section><\/section>
                        \r\n

                        \u00a0Contingency Tables \u2014 Homework<\/h2>\r\n101. 0\r\n\r\n103. [latex]\\frac{{10}}{{67}}[\/latex]\r\n\r\n105.[latex]\\frac{{10}}{{34}}[\/latex]\r\n\r\n107. d\r\n\r\n109.\r\n\r\na.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                        Race and Sex<\/th>\r\n1\u201314<\/th>\r\n15\u201324<\/th>\r\n25\u201364<\/th>\r\nover 64<\/th>\r\nTOTAL<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                        white, male<\/td>\r\n210<\/td>\r\n3,360<\/td>\r\n13,610<\/td>\r\n4,870<\/td>\r\n22,050<\/td>\r\n<\/tr>\r\n
                        white, female<\/td>\r\n80<\/td>\r\n580<\/td>\r\n3,380<\/td>\r\n890<\/td>\r\n4,930<\/td>\r\n<\/tr>\r\n
                        black, male<\/td>\r\n10<\/td>\r\n460<\/td>\r\n1,060<\/td>\r\n140<\/td>\r\n1,670<\/td>\r\n<\/tr>\r\n
                        black, female<\/td>\r\n0<\/td>\r\n40<\/td>\r\n270<\/td>\r\n20<\/td>\r\n330<\/td>\r\n<\/tr>\r\n
                        all others<\/td>\r\n100<\/td>\r\n<\/tr>\r\n
                        TOTALS<\/td>\r\n310<\/td>\r\n4,650<\/td>\r\n18,780<\/td>\r\n6,020<\/td>\r\n29,760<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nb.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                        Race and Sex<\/th>\r\n1\u201314<\/th>\r\n15\u201324<\/th>\r\n25\u201364<\/th>\r\nover 64<\/th>\r\nTOTAL<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                        white, male<\/td>\r\n210<\/td>\r\n3,360<\/td>\r\n13,610<\/td>\r\n4,870<\/td>\r\n22,050<\/td>\r\n<\/tr>\r\n
                        white, female<\/td>\r\n80<\/td>\r\n580<\/td>\r\n3,380<\/td>\r\n890<\/td>\r\n4,930<\/td>\r\n<\/tr>\r\n
                        black, male<\/td>\r\n10<\/td>\r\n460<\/td>\r\n1,060<\/td>\r\n140<\/td>\r\n1,670<\/td>\r\n<\/tr>\r\n
                        black, female<\/td>\r\n0<\/td>\r\n40<\/td>\r\n270<\/td>\r\n20<\/td>\r\n330<\/td>\r\n<\/tr>\r\n
                        all others<\/td>\r\n10<\/td>\r\n210<\/td>\r\n460<\/td>\r\n100<\/td>\r\n780<\/td>\r\n<\/tr>\r\n
                        TOTALS<\/td>\r\n310<\/td>\r\n4,650<\/td>\r\n18,780<\/td>\r\n6,020<\/td>\r\n29,760<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nc. [latex]\\frac{{22050}}{{29760}}[\/latex]\r\n\r\nd. [latex]\\frac{{330}}{{29760}}[\/latex]\r\n\r\ne. [latex]\\frac{{2000}}{{29760}}[\/latex]\r\n\r\nf. [latex]\\frac{{23720}}{{29760}}[\/latex]\r\n\r\ng. [latex]\\frac{{5010}}{{6020}}[\/latex]\r\n\r\n<\/section>111. b\r\n\r\n113.\r\n\r\na. [latex]\\frac{{26}}{{106}}[\/latex]\r\n\r\nb. [latex]\\frac{{33}}{{106}}[\/latex]\r\n\r\nc. [latex]\\frac{{21}}{{106}}[\/latex]\r\n\r\nd. [latex]\\left(\\frac{{26}}{{106}}\\right)+\\left(\\frac{{33}}{{106}}\\right)-\\left(\\frac{{21}}{{106}}\\right)=\\left(\\frac{{38}}{{106}}\\right)[\/latex]\r\n\r\ne. [latex]\\frac{{21}}{{33}}[\/latex]\r\n

                        Tree and Venn Diagrams \u2014 Homework<\/h2>\r\n115. a\r\n
                        \r\n

                        Extra Practice<\/h2>\r\n<\/div>\r\n118.\r\n\r\n \r\n\r\na. P<\/em>(C<\/em>) = 0.4567\r\n\r\nb. not enough information\r\n\r\nc. not enough information\r\n\r\nd. No, because over half (0.51) of men have at least one false positive text\r\n\r\n120.\r\n\r\n \r\n
                          \r\n \t
                        1. P<\/em>(J<\/em>\u00a0OR\u00a0K<\/em>) =\u00a0P<\/em>(J<\/em>) +\u00a0P<\/em>(K<\/em>) \u2212\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>); 0.45 = 0.18 + 0.37 -\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>); solve to find\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>) = 0.10<\/li>\r\n \t
                        2. P<\/em>(NOT (J<\/em>\u00a0AND\u00a0K<\/em>)) = 1 -\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>) = 1 - 0.10 = 0.90<\/li>\r\n \t
                        3. P<\/em>(NOT (J<\/em>\u00a0OR\u00a0K<\/em>)) = 1 -\u00a0P<\/em>(J<\/em>\u00a0OR\u00a0K<\/em>) = 1 - 0.45 = 0.55<\/li>\r\n<\/ol>\r\n \r\n\r\n \r\n\r\n121.\r\n\r\na.\r\n\r\n
                          \r\n
                          \r\n
                          \"This\r\n
                          b. P<\/em>(GG<\/em>) =[latex]\\left(\\frac{{5}}{{8}}\\right)\\left(\\frac{{5}}{{8}}\\right)=\\frac{{25}}{{64}}[\/latex]<\/section>
                          c. P<\/em>(at least one green) = P<\/em>(GG<\/em>) + P<\/em>(GY<\/em>) + P<\/em>(YG<\/em>) =\u00a0[latex]\\left(\\frac{{25}}{{64}}\\right)+\\left(\\frac{{15}}{{64}}\\right)+\\left(\\frac{{15}}{{64}}\\right)=\\frac{{55}}{{64}}[\/latex]<\/section>
                          d. P<\/em>(G<\/em>|G<\/em>) = 58.\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/section>
                          e. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.<\/span>123.\r\n
                          \r\n
                          a.<\/div>\r\n
                          \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                          <\/th>\r\n<20<\/th>\r\n20\u201364<\/th>\r\n>64<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                          Female<\/strong><\/td>\r\n0.0244<\/td>\r\n0.3954<\/td>\r\n0.0661<\/td>\r\n0.486<\/td>\r\n<\/tr>\r\n
                          Male<\/strong><\/td>\r\n0.0259<\/td>\r\n0.4186<\/td>\r\n0.0695<\/td>\r\n0.514<\/td>\r\n<\/tr>\r\n
                          Totals<\/strong><\/td>\r\n0.0503<\/td>\r\n0.8140<\/td>\r\n0.1356<\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nb. P<\/em>(F<\/em>) = 0.486\r\n\r\nc. P<\/em>(>64|F<\/em>) = 0.1361\r\n\r\nd. P<\/em>(>64 and F<\/em>) = P<\/em>(F<\/em>) P<\/em>(>64|F<\/em>) = (0.486)(0.1361) = 0.0661\r\n\r\ne. P<\/em>(>64|F<\/em>) is the percentage of female drivers who are 65 or older and P<\/em>(>64 and F<\/em>) is the percentage of drivers who are female and 65 or older.\r\n\r\nf. P<\/em>(>64<\/em>) = P<\/em>(>64 and F<\/em>) + P<\/em>(>64 and M<\/em>) = 0.1356\r\n\r\ng. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 andF<\/em>) = 0.0661.\r\n\r\n125.\r\n
                            \r\n \t
                          1. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                            Car, Truck or Van<\/th>\r\nWalk<\/th>\r\nPublic Transportation<\/th>\r\nOther<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                            Alone<\/strong><\/td>\r\n0.7318<\/td>\r\n<\/tr>\r\n
                            Not Alone<\/strong><\/td>\r\n0.1332<\/td>\r\n<\/tr>\r\n
                            Totals<\/strong><\/td>\r\n0.8650<\/td>\r\n0.0390<\/td>\r\n0.0530<\/td>\r\n0.0430<\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t
                          2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P<\/em>(Alone) = 0.7318 + 0.0390 = 0.7708.<\/li>\r\n \t
                          3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771<\/li>\r\n \t
                          4. (0.1332)(1,000) = 133<\/li>\r\n<\/ol>\r\n127.\r\n
                            \r\n
                            \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                            Homosexual\/Bisexual<\/th>\r\nIV Drug User*<\/th>\r\nHeterosexual Contact<\/th>\r\nOther<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                            Female<\/td>\r\n0<\/td>\r\n70<\/td>\r\n136<\/td>\r\n49<\/td>\r\n255<\/strong><\/td>\r\n<\/tr>\r\n
                            Male<\/td>\r\n2,146<\/td>\r\n463<\/td>\r\n60<\/td>\r\n135<\/td>\r\n2,804<\/strong><\/td>\r\n<\/tr>\r\n
                            Totals<\/td>\r\n2,146<\/strong><\/td>\r\n533<\/strong><\/td>\r\n196<\/strong><\/td>\r\n184<\/strong><\/td>\r\n3,059<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                              \r\n \t
                            1. [latex]\\frac{{255}}{{3059}}[\/latex]<\/li>\r\n \t
                            2. [latex]\\frac{{196}}{{3059}}[\/latex]<\/li>\r\n \t
                            3. [latex]\\frac{{718}}{{3059}}[\/latex]<\/li>\r\n \t
                            4. 0<\/li>\r\n \t
                            5. [latex]\\frac{{463}}{{3059}}[\/latex]<\/li>\r\n \t
                            6. [latex]\\frac{{136}}{{196}}[\/latex]<\/li>\r\n \t
                            7. \r\n
                              \"\"<\/span><\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                              <\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                              <\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section> ","rendered":"

                              1.<\/p>\n

                              \n
                                \n
                              1. P<\/em>(L\u2032<\/em>) = P<\/em>(S<\/em>)<\/li>\n
                              2. P<\/em>(M<\/em> OR S<\/em>)<\/li>\n
                              3. P<\/em>(F<\/em> AND L<\/em>)<\/li>\n
                              4. P<\/em>(M<\/em>|L<\/em>)<\/li>\n
                              5. P<\/em>(L<\/em>|M<\/em>)<\/li>\n
                              6. P<\/em>(S<\/em>|F<\/em>)<\/li>\n
                              7. P<\/em>(F<\/em>|L<\/em>)<\/li>\n
                              8. P<\/em>(F<\/em> OR L<\/em>)<\/li>\n
                              9. P<\/em>(M<\/em> AND S<\/em>)<\/li>\n
                              10. P<\/em>(F<\/em>)<\/li>\n<\/ol>\n

                                3.\u00a0P<\/em>(N<\/em>) = [latex]\\frac{{15}}{{42}}=\\frac{{5}}{{14}}=0.36[\/latex]<\/p>\n

                                5.\u00a0P<\/em>(C<\/em>) = [latex]\\frac{{5}}{{42}}=0.12[\/latex]<\/p>\n

                                7.\u00a0P<\/em>(G<\/em>) = [latex]\\frac{{20}}{{150}}=\\frac{{2}}{{15}}=0.13[\/latex]<\/p>\n

                                9.\u00a0P<\/em>(R<\/em>) = [latex]\\frac{{22}}{{150}}=\\frac{{11}}{{75}}=0.15[\/latex]<\/p>\n

                                11.\u00a0P<\/em>(O<\/em>) = [latex]\\frac{{{22}-{38}-{20}-{28}-{26}}}{{150}}=\\frac{{16}}{{150}}=\\frac{{8}}{{75}}=0.11[\/latex]<\/p>\n

                                13.\u00a0P<\/em>(E<\/em>) = [latex]\\frac{{47}}{{194}}=0.24[\/latex]<\/p>\n

                                15.\u00a0P<\/em>(N<\/em>) = [latex]\\frac{{23}}{{194}}=0.12[\/latex]<\/p>\n

                                17.\u00a0P<\/em>(S<\/em>)=[latex]\\frac{{12}}{{194}}={{6}}{{97}}=0.06[\/latex]<\/p>\n

                                19. [latex]\\frac{{13}}{{52}}={{1}}{{4}}=0.25[\/latex]<\/p>\n

                                21. [latex]\\frac{{3}}{{6}}={{1}}{{2}}=0.5[\/latex]<\/p>\n

                                23. P(R) = [latex]\\frac{{4}}{{8}}=0.5[\/latex]<\/p>\n

                                25. P(O or H)<\/p>\n

                                27. P(H | I)<\/p>\n

                                29. P(N | O)<\/p>\n

                                31. P(I or N)<\/p>\n

                                33. P(I)<\/p>\n

                                35.\u00a0The likelihood that an event will occur given that another event has already occurred.<\/p>\n

                                37. 1<\/p>\n

                                39.\u00a0the probability of landing on an even number or a multiple of three<\/p>\n<\/section>\n

                                Independent and Mutually Exclusive Events \u2014 Practice<\/h2>\n

                                41. P(J) = 0.3
                                \n43. P(Q AND R) = P(Q)P(R)0.1 = (0.4)P(R)P(R) = 0.25<\/p>\n

                                Two Basic Rules of Probability \u2014 Practice<\/h2>\n

                                45. 0.376
                                \n47. C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.
                                \n49. L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.
                                \n51. 0.6492
                                \n53. No, because P(L AND C) does not equal 0.<\/p>\n

                                \u00a0Contingency Tables \u2014 Practice<\/h2>\n

                                55. P(musician is a male AND had private instruction) = [latex]\\frac{15}{130} = [latex]\\frac{3}{26} = 0.12[\/latex]
                                \n57. The events are not mutually exclusive. It is possible to be a female musician who learned music in school.<\/p>\n

                                \n

                                Tree and Venn Diagrams \u2014 Practice<\/h2>\n<\/div>\n

                                58.\"This<\/p>\n

                                60. [latex]\\frac{35,065}{100,450}[\/latex]<\/p>\n

                                62. To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is [latex]\\frac{4,715}{100,450}[\/latex].<\/p>\n

                                64. To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is [latex]\\frac{4715}{15,273}[\/latex].<\/p>\n

                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n
                                \n

                                Terminology \u2014 Homework<\/h2>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n

                                67.<\/p>\n

                                  \n
                                1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%<\/li>\n
                                2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.<\/li>\n<\/ol>\n
                                  \n
                                  \n
                                  \n

                                  Independent and Mutually Exclusive Events \u2014 Homework<\/h2>\n<\/div>\n<\/section>\n<\/div>\n
                                  <\/section>\n

                                  69. 0
                                  \n71. 0.3571
                                  \n73. 0.2142
                                  \n75. Physician (83.7)
                                  \n77. 83.7 \u2212 79.6 = 4.1
                                  \n79. P(Occupation < 81.3) = 0.5<\/p>\n

                                  \n
                                  \n
                                  \n

                                  Two Basic Rules of Probability \u2014 Homework<\/h2>\n<\/div>\n<\/section>\n<\/div>\n

                                  81.<\/p>\n

                                  \n
                                    \n
                                  1. The Forum Research surveyed 1,046 Torontonians.<\/li>\n
                                  2. 58%<\/li>\n
                                  3. 42% of 1,046 = 439 (rounding to the nearest integer)<\/li>\n
                                  4. 0.57<\/li>\n
                                  5. 0.60.<\/li>\n<\/ol>\n

                                    83.<\/p>\n

                                      \n
                                    1. P<\/em>(Betting on two line that touch each other on the table) = [latex]\\frac{{6}}{{38}}[\/latex]<\/li>\n
                                    2. P<\/em>(Betting on three numbers in a line) = [latex]\\frac{{3}}{{38}}[\/latex]<\/li>\n
                                    3. P<\/em>(Bettting on one number) = [latex]\\frac{{1}}{{38}}[\/latex]<\/li>\n
                                    4. P<\/em>(Betting on four number that touch each other to form a square) = [latex]\\frac{{4}}{{38}}[\/latex]<\/li>\n
                                    5. P<\/em>(Betting on two number that touch each other on the table ) = [latex]\\frac{{2}}{{38}}[\/latex]<\/li>\n
                                    6. P<\/em>(Betting on 0-00-1-2-3) = [latex]\\frac{{5}}{{38}}[\/latex]<\/li>\n
                                    7. P<\/em>(Betting on 0-1-2; or 0-00-2; or 00-2-3) = [latex]\\frac{{3}}{{38}}[\/latex]<\/li>\n<\/ol>\n<\/section>\n

                                      85.<\/p>\n

                                      \n
                                        \n
                                      1. {G<\/em>1, G<\/em>2, G<\/em>3, G<\/em>4, G<\/em>5, Y<\/em>1, Y<\/em>2, Y<\/em>3}<\/li>\n
                                      2. [latex]\\frac{{5}}{{8}}[\/latex]<\/li>\n
                                      3. [latex]\\frac{{2}}{{3}}[\/latex]<\/li>\n
                                      4. [latex]\\frac{{2}}{{8}}[\/latex]<\/li>\n
                                      5. [latex]\\frac{{6}}{{8}}[\/latex]<\/li>\n
                                      6. No, because P<\/em>(G<\/em> AND E<\/em>) does not equal 0.<\/li>\n<\/ol>\n

                                        87.<\/p>\n

                                        \n
                                        \n
                                        NOTE<\/div>\n<\/header>\n
                                        \n

                                        The coin toss is independent of the card picked first.<\/p>\n<\/section>\n<\/div>\n

                                          \n
                                        1. {(G<\/em>,H<\/em>) (G<\/em>,T<\/em>) (B<\/em>,H<\/em>) (B<\/em>,T<\/em>) (R<\/em>,H<\/em>) (R<\/em>,T<\/em>)}<\/li>\n
                                        2. P<\/em>(A<\/em>) = P<\/em>(blue)P<\/em>(head) = ([latex]\\frac{{3}}{{10}}[\/latex])([latex]\\frac{{1}}{{2}}[\/latex])=[latex]\\frac{{3}}{{20}}[\/latex]<\/li>\n
                                        3. Yes, A<\/em> and B<\/em> are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P<\/em>(A<\/em> AND B<\/em>) = 0<\/li>\n
                                        4. No, A<\/em> and C<\/em> are not mutually exclusive because they can occur at the same time. In fact, C<\/em> includes all of the outcomes of A<\/em>; if the card chosen is blue it is also (red or blue). P<\/em>(A<\/em> AND C<\/em>) = P<\/em>(A<\/em>) = 3<\/span>20<\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n<\/ol>\n

                                          89.<\/p>\n

                                          \n
                                            \n
                                          1. S<\/em> = {(HHH<\/em>), (HHT<\/em>), (HTH<\/em>), (HTT<\/em>), (THH<\/em>), (THT<\/em>), (TTH<\/em>), (TTT<\/em>)}<\/li>\n
                                          2. [latex]\\frac{{4}}{{8}}[\/latex]<\/li>\n
                                          3. Yes, because if A<\/em> has occurred, it is impossible to obtain two tails. In other words, P<\/em>(A<\/em> AND B<\/em>) = 0.<\/li>\n<\/ol>\n

                                            91.<\/p>\n

                                            \n
                                              \n
                                            1. If Y<\/em> and Z<\/em> are independent, then P<\/em>(Y<\/em> AND Z<\/em>) = P<\/em>(Y<\/em>)P<\/em>(Z<\/em>), so P<\/em>(Y<\/em> OR Z<\/em>) = P<\/em>(Y<\/em>) + P<\/em>(Z<\/em>) - P<\/em>(Y<\/em>)P<\/em>(Z<\/em>).<\/li>\n
                                            2. 0.5<\/li>\n<\/ol>\n

                                              93.<\/p>\n

                                              \n

                                              1. iii<\/span>\u00a02.\u00a0i<\/span>\u00a03.\u00a0iv<\/span>\u00a04.\u00a0ii<\/span><\/span><\/p>\n<\/section>\n

                                              95.<\/p>\n

                                              \n
                                                \n
                                              1. P<\/em>(R<\/em>) = 0.44<\/li>\n
                                              2. P<\/em>(R<\/em>|E<\/em>) = 0.56<\/li>\n
                                              3. P<\/em>(R<\/em>|O<\/em>) = 0.31<\/li>\n
                                              4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P<\/em>(R<\/em>|E<\/em>) \u2260 P<\/em>(R<\/em>).<\/li>\n
                                              5. No, this study definitely does not support that notion; in fact<\/u><\/em>, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P<\/em>(R<\/em>|E<\/em>) > P<\/em>(R<\/em>).<\/li>\n<\/ol>\n<\/section>\n
                                                \u00a097.<\/p>\n
                                                \n
                                                  \n
                                                1. \n

                                                  P<\/em>(type O OR Rh-) = P<\/em>(type O) + P<\/em>(Rh-) - P<\/em>(type O AND Rh-)<\/p>\n

                                                  0.52 = 0.43 + 0.15 - P<\/em>(type O AND Rh-); solve to find P<\/em>(type O AND Rh-) = 0.06<\/p>\n

                                                  6% of people have type O, Rh- blood<\/p>\n<\/li>\n

                                                2. \n

                                                  P<\/em>(NOT(type O AND Rh-)) = 1 - P<\/em>(type O AND Rh-) = 1 - 0.06 = 0.94<\/p>\n

                                                  94% of people do not have type O, Rh- blood<\/p>\n<\/li>\n<\/ol>\n

                                                  99.<\/p>\n

                                                  \n
                                                    \n
                                                  1. Let C<\/em> = be the event that the cookie contains chocolate. Let N<\/em> = the event that the cookie contains nuts.<\/li>\n
                                                  2. P<\/em>(C<\/em> OR N<\/em>) = P<\/em>(C<\/em>) + P<\/em>(N<\/em>) - P<\/em>(C<\/em> AND N<\/em>) = 0.36 + 0.12 - 0.08 = 0.40<\/li>\n
                                                  3. P<\/em>(NEITHER chocolate NOR nuts) = 1 - P<\/em>(C<\/em> OR N<\/em>) = 1 - 0.40 = 0.60<\/li>\n<\/ol>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n
                                                    \n

                                                    \u00a0Contingency Tables \u2014 Homework<\/h2>\n

                                                    101. 0<\/p>\n

                                                    103. [latex]\\frac{{10}}{{67}}[\/latex]<\/p>\n

                                                    105.[latex]\\frac{{10}}{{34}}[\/latex]<\/p>\n

                                                    107. d<\/p>\n

                                                    109.<\/p>\n

                                                    a.<\/p>\n\n\n\n\n\n\n\n\n\n\n
                                                    Race and Sex<\/th>\n1\u201314<\/th>\n15\u201324<\/th>\n25\u201364<\/th>\nover 64<\/th>\nTOTAL<\/th>\n<\/tr>\n<\/thead>\n
                                                    white, male<\/td>\n210<\/td>\n3,360<\/td>\n13,610<\/td>\n4,870<\/td>\n22,050<\/td>\n<\/tr>\n
                                                    white, female<\/td>\n80<\/td>\n580<\/td>\n3,380<\/td>\n890<\/td>\n4,930<\/td>\n<\/tr>\n
                                                    black, male<\/td>\n10<\/td>\n460<\/td>\n1,060<\/td>\n140<\/td>\n1,670<\/td>\n<\/tr>\n
                                                    black, female<\/td>\n0<\/td>\n40<\/td>\n270<\/td>\n20<\/td>\n330<\/td>\n<\/tr>\n
                                                    all others<\/td>\n100<\/td>\n<\/tr>\n
                                                    TOTALS<\/td>\n310<\/td>\n4,650<\/td>\n18,780<\/td>\n6,020<\/td>\n29,760<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                    b.<\/p>\n\n\n\n\n\n\n\n\n\n\n
                                                    Race and Sex<\/th>\n1\u201314<\/th>\n15\u201324<\/th>\n25\u201364<\/th>\nover 64<\/th>\nTOTAL<\/th>\n<\/tr>\n<\/thead>\n
                                                    white, male<\/td>\n210<\/td>\n3,360<\/td>\n13,610<\/td>\n4,870<\/td>\n22,050<\/td>\n<\/tr>\n
                                                    white, female<\/td>\n80<\/td>\n580<\/td>\n3,380<\/td>\n890<\/td>\n4,930<\/td>\n<\/tr>\n
                                                    black, male<\/td>\n10<\/td>\n460<\/td>\n1,060<\/td>\n140<\/td>\n1,670<\/td>\n<\/tr>\n
                                                    black, female<\/td>\n0<\/td>\n40<\/td>\n270<\/td>\n20<\/td>\n330<\/td>\n<\/tr>\n
                                                    all others<\/td>\n10<\/td>\n210<\/td>\n460<\/td>\n100<\/td>\n780<\/td>\n<\/tr>\n
                                                    TOTALS<\/td>\n310<\/td>\n4,650<\/td>\n18,780<\/td>\n6,020<\/td>\n29,760<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                    c. [latex]\\frac{{22050}}{{29760}}[\/latex]<\/p>\n

                                                    d. [latex]\\frac{{330}}{{29760}}[\/latex]<\/p>\n

                                                    e. [latex]\\frac{{2000}}{{29760}}[\/latex]<\/p>\n

                                                    f. [latex]\\frac{{23720}}{{29760}}[\/latex]<\/p>\n

                                                    g. [latex]\\frac{{5010}}{{6020}}[\/latex]<\/p>\n<\/section>\n

                                                    111. b<\/p>\n

                                                    113.<\/p>\n

                                                    a. [latex]\\frac{{26}}{{106}}[\/latex]<\/p>\n

                                                    b. [latex]\\frac{{33}}{{106}}[\/latex]<\/p>\n

                                                    c. [latex]\\frac{{21}}{{106}}[\/latex]<\/p>\n

                                                    d. [latex]\\left(\\frac{{26}}{{106}}\\right)+\\left(\\frac{{33}}{{106}}\\right)-\\left(\\frac{{21}}{{106}}\\right)=\\left(\\frac{{38}}{{106}}\\right)[\/latex]<\/p>\n

                                                    e. [latex]\\frac{{21}}{{33}}[\/latex]<\/p>\n

                                                    Tree and Venn Diagrams \u2014 Homework<\/h2>\n

                                                    115. a<\/p>\n

                                                    \n

                                                    Extra Practice<\/h2>\n<\/div>\n

                                                    118.<\/p>\n

                                                     <\/p>\n

                                                    a. P<\/em>(C<\/em>) = 0.4567<\/p>\n

                                                    b. not enough information<\/p>\n

                                                    c. not enough information<\/p>\n

                                                    d. No, because over half (0.51) of men have at least one false positive text<\/p>\n

                                                    120.<\/p>\n

                                                     <\/p>\n

                                                      \n
                                                    1. P<\/em>(J<\/em>\u00a0OR\u00a0K<\/em>) =\u00a0P<\/em>(J<\/em>) +\u00a0P<\/em>(K<\/em>) \u2212\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>); 0.45 = 0.18 + 0.37 -\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>); solve to find\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>) = 0.10<\/li>\n
                                                    2. P<\/em>(NOT (J<\/em>\u00a0AND\u00a0K<\/em>)) = 1 -\u00a0P<\/em>(J<\/em>\u00a0AND\u00a0K<\/em>) = 1 - 0.10 = 0.90<\/li>\n
                                                    3. P<\/em>(NOT (J<\/em>\u00a0OR\u00a0K<\/em>)) = 1 -\u00a0P<\/em>(J<\/em>\u00a0OR\u00a0K<\/em>) = 1 - 0.45 = 0.55<\/li>\n<\/ol>\n

                                                       <\/p>\n

                                                       <\/p>\n

                                                      121.<\/p>\n

                                                      a.<\/p>\n

                                                      \n
                                                      \n
                                                      \n
                                                      \n
                                                      \"This<\/p>\n
                                                      \n
                                                      b. P<\/em>(GG<\/em>) =[latex]\\left(\\frac{{5}}{{8}}\\right)\\left(\\frac{{5}}{{8}}\\right)=\\frac{{25}}{{64}}[\/latex]<\/section>\n
                                                      c. P<\/em>(at least one green) = P<\/em>(GG<\/em>) + P<\/em>(GY<\/em>) + P<\/em>(YG<\/em>) =\u00a0[latex]\\left(\\frac{{25}}{{64}}\\right)+\\left(\\frac{{15}}{{64}}\\right)+\\left(\\frac{{15}}{{64}}\\right)=\\frac{{55}}{{64}}[\/latex]<\/section>\n
                                                      d. P<\/em>(G<\/em>|G<\/em>) = 58.\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/section>\n
                                                      e. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.<\/span>123.<\/p>\n
                                                      \n
                                                      \n
                                                      a.<\/div>\n
                                                      \n
                                                      \n\n\n\n\n\n\n\n
                                                      <\/th>\n<20<\/th>\n20\u201364<\/th>\n>64<\/th>\nTotal<\/th>\n<\/tr>\n<\/thead>\n
                                                      Female<\/strong><\/td>\n0.0244<\/td>\n0.3954<\/td>\n0.0661<\/td>\n0.486<\/td>\n<\/tr>\n
                                                      Male<\/strong><\/td>\n0.0259<\/td>\n0.4186<\/td>\n0.0695<\/td>\n0.514<\/td>\n<\/tr>\n
                                                      Totals<\/strong><\/td>\n0.0503<\/td>\n0.8140<\/td>\n0.1356<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                      b. P<\/em>(F<\/em>) = 0.486<\/p>\n

                                                      c. P<\/em>(>64|F<\/em>) = 0.1361<\/p>\n

                                                      d. P<\/em>(>64 and F<\/em>) = P<\/em>(F<\/em>) P<\/em>(>64|F<\/em>) = (0.486)(0.1361) = 0.0661<\/p>\n

                                                      e. P<\/em>(>64|F<\/em>) is the percentage of female drivers who are 65 or older and P<\/em>(>64 and F<\/em>) is the percentage of drivers who are female and 65 or older.<\/p>\n

                                                      f. P<\/em>(>64<\/em>) = P<\/em>(>64 and F<\/em>) + P<\/em>(>64 and M<\/em>) = 0.1356<\/p>\n

                                                      g. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 andF<\/em>) = 0.0661.<\/p>\n

                                                      125.<\/p>\n

                                                        \n
                                                      1. \n\n\n\n\n\n\n\n
                                                        Car, Truck or Van<\/th>\nWalk<\/th>\nPublic Transportation<\/th>\nOther<\/th>\nTotal<\/th>\n<\/tr>\n<\/thead>\n
                                                        Alone<\/strong><\/td>\n0.7318<\/td>\n<\/tr>\n
                                                        Not Alone<\/strong><\/td>\n0.1332<\/td>\n<\/tr>\n
                                                        Totals<\/strong><\/td>\n0.8650<\/td>\n0.0390<\/td>\n0.0530<\/td>\n0.0430<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n
                                                      2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P<\/em>(Alone) = 0.7318 + 0.0390 = 0.7708.<\/li>\n
                                                      3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771<\/li>\n
                                                      4. (0.1332)(1,000) = 133<\/li>\n<\/ol>\n

                                                        127.<\/p>\n

                                                        \n
                                                        \n
                                                        \n
                                                        \n\n\n\n\n\n\n\n
                                                        Homosexual\/Bisexual<\/th>\nIV Drug User*<\/th>\nHeterosexual Contact<\/th>\nOther<\/th>\nTotal<\/th>\n<\/tr>\n<\/thead>\n
                                                        Female<\/td>\n0<\/td>\n70<\/td>\n136<\/td>\n49<\/td>\n255<\/strong><\/td>\n<\/tr>\n
                                                        Male<\/td>\n2,146<\/td>\n463<\/td>\n60<\/td>\n135<\/td>\n2,804<\/strong><\/td>\n<\/tr>\n
                                                        Totals<\/td>\n2,146<\/strong><\/td>\n533<\/strong><\/td>\n196<\/strong><\/td>\n184<\/strong><\/td>\n3,059<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                          \n
                                                        1. [latex]\\frac{{255}}{{3059}}[\/latex]<\/li>\n
                                                        2. [latex]\\frac{{196}}{{3059}}[\/latex]<\/li>\n
                                                        3. [latex]\\frac{{718}}{{3059}}[\/latex]<\/li>\n
                                                        4. 0<\/li>\n
                                                        5. [latex]\\frac{{463}}{{3059}}[\/latex]<\/li>\n
                                                        6. [latex]\\frac{{136}}{{196}}[\/latex]<\/li>\n
                                                        7. \n
                                                          \"\"<\/span><\/figure>\n<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n
                                                          <\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n
                                                          <\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n

                                                           <\/p>\n\n\t\t\t

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                                                          Candela Citations<\/h3>\n\t\t\t\t\t
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