{"id":89,"date":"2019-06-18T12:52:18","date_gmt":"2019-06-18T16:52:18","guid":{"rendered":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/more-probability\/"},"modified":"2023-12-17T19:12:25","modified_gmt":"2023-12-18T00:12:25","slug":"more-probability","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/more-probability\/","title":{"raw":"7.1. Binomial Probability","rendered":"7.1. Binomial Probability"},"content":{"raw":"
[Latexpage]<\/div>\r\n

Binomial Probability<\/h1>\r\nIn this section, we will consider types of problems that involve a sequence of trials, where each trial has only two outcomes, a success or a failure. These trials are independent. That is, the outcome of one does not affect the outcome of any other trial. Furthermore, the probability of success, p<\/em>, and the probability of failure, (1 \u2212 p<\/em>), remains the same throughout the experiment. These problems are called binomial probability<\/strong> problems. Since these problems were researched by a Swiss mathematician named Jacques Bernoulli around 1700, they are also referred to as Bernoulli trials<\/strong>.\r\n\r\nWe give the following definition:\r\n
\r\n\r\nBinomial Experiment<\/strong>: A binomial experiment satisfies the following four conditions:\r\n
    \r\n \t
  1. There are only two outcomes, a success or a failure, for each trial.<\/li>\r\n \t
  2. The same experiment is repeated several times.<\/li>\r\n \t
  3. The trials are independent; that is, the outcome of a particular trial does not affect the outcome of any other trial.<\/li>\r\n \t
  4. The probability of success remains the same for every trial.<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\nThe probability model that we are about to investigate will give us the tools to solve many real-life problems like the ones given below.\r\n
      \r\n \t
    1. If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?<\/li>\r\n \t
    2. If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 7 out of 10 free throws in a game?<\/li>\r\n \t
    3. If a medicine cures 80% of the people who take it, what is the probability that among the 10 people who take the medicine, 6 will be cured?<\/li>\r\n \t
    4. If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?<\/li>\r\n \t
    5. If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?<\/li>\r\n<\/ol>\r\nWe now consider the following example to develop a formula for finding the probability of k<\/em> successes in n<\/em> Bernoulli trials.\r\n\r\n \r\n
      \r\n

      Example 7.1.1<\/p>\r\n\r\n<\/header>\r\n

      A baseball player has a batting average of 0.300. If he bats four times in a game, find the probability that he will have:<\/div>\r\n
      a.<\/strong> four hits<\/div>\r\n
      b.<\/strong> three hits<\/div>\r\n
      c.<\/strong> two hits<\/div>\r\n
      d.<\/strong> one hit<\/div>\r\n
      e.<\/strong> no hits.<\/div>\r\n
      Solution <\/strong><\/div>\r\n
      Let us suppose S<\/em> denotes that the player gets a hit, and F<\/em> denotes that he does not get a hit. This is a binomial experiment because it meets all four conditions. First, there are only two outcomes, S<\/em> or F<\/em>. Clearly the experiment is repeated four times. Lastly, if we assume that the player's skillfulness to get a hit does not change each time he comes to bat, the trials are independent with a probability of 0.3 of getting a hit during each trial. We draw a tree diagram to show all situations.<\/div>\r\n
      \r\n
      \\begin{tikzpicture}[grow=right, sloped]\r\n\\tikzstyle{level 1}=[level distance=3cm, sibling distance=8cm]\r\n\\tikzstyle{level 2}=[level distance=3cm, sibling distance=4cm]\r\n\\tikzstyle{level 3}=[level distance=3cm, sibling distance=2cm]\r\n\\tikzstyle{level 4}=[level distance=2cm, sibling distance=1cm]\r\n\\node {}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad FFFF=(0.7)^4$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad FFFS=(0.3)(0.7)^3$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad FFSF=(0.3)(0.7)^3$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad FFSS=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad FSFF=(0.3)(0.7)^3$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad FSFS=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad FSSF=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad FSSS=(0.3)^3(0.7)$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad SFFF=(0.3)(0.7)^3$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad SFFS=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad SFSF=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad SFSS=(0.3)^3(0.7)$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode {$F$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad SSFF=(0.3)^2(0.7)^2$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad SSFS=(0.3)^3(0.7)$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode {$S$}\r\nchild {\r\nnode[label=right:\r\n{$F\\quad SSSF=(0.3)^3(0.7)$}] {}\r\nedge from parent\r\nnode[below] {$0.7$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$S\\quad SSSS=(0.3)^4$}] {}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n}\r\nedge from parent\r\nnode[above] {$0.3$}\r\n};\r\n\\end{tikzpicture}<\/div>\r\n<\/div>\r\n
      Let us first find the probability of getting, for example, two hits. We will have to consider the six possibilities, SSFF , SFSF , SFFS , FSSF , FSFS , FFSS , as shown in the above tree diagram. We list the probabilities of each below.<\/div>\r\n
      P<\/em>(SSFF) = (0.3)(0.3)(0.7)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\r\n
      P<\/em>(SFSF) = (0.3)(0.7)(0.3)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\r\n
      P<\/em>(SFFS) = (0.3)(0.7)(0.7)(0.3) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\r\n
      P<\/em>(FSSF) = (0.7)(0.3)(0.3)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\r\n
      P<\/em>(FSFS) = (0.7)(0.3)(0.7)(0.3) = (0.3)2<\/sup>(0.7)2<\/sup><\/div>\r\n
      P<\/em>(FFSS) = (0.7)(0.7)(0.3)(0.3) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\r\n
      Since the probability of each of these six outcomes is (0.3)2<\/sup>(0.7)2<\/sup>, the probability of obtaining two successes is 6(0.3)2<\/sup>(0.7)2<\/sup>.<\/div>\r\n
      The probability of getting one hit can be obtained in the same way. Since each permutation has one S<\/em> and three F<\/em>'s, there are four such outcomes: SFFF , FSFF , FFSF , and FFFS.<\/div>\r\n
      And since the probability of each of the four outcomes is (0.3)(0.7)3<\/sup>, the probability of getting one hit is 4(0.3)(0.7)3<\/sup>.<\/div>\r\n
      The table below lists the probabilities for all cases, and shows a comparison with the binomial expansion of fourth degree. Again, p<\/em> denotes the probability of success, and q<\/em> = (1 \u2212 p<\/em>) the probability of failure.<\/div>\r\n
      \r\n\r\n\r\n\r\n\r\n
      Outcome<\/td>\r\nFour Hits<\/td>\r\nThree hits<\/td>\r\nTwo Hits<\/td>\r\nOne hits<\/td>\r\nNo Hits<\/td>\r\n<\/tr>\r\n
      Probability<\/td>\r\n(0.3)4<\/sup><\/td>\r\n4(0.3)3<\/sup>(0.7)<\/td>\r\n6(0.3)2<\/sup>(0.7)2<\/sup><\/td>\r\n4(0.3)(0.7)3<\/sup><\/td>\r\n(0.7)4<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nThis gives us the following theorem:\r\n
      \r\n\r\nBinomial Probability Theorem<\/strong>:\r\n\r\nThe probability of obtaining k<\/em> successes in n<\/em> independent Bernoulli trials is given by:\r\n

      P<\/em>(n<\/em>, k<\/em>; p<\/em>) = n<\/em>Ckp<\/em>k<\/sup>q<\/em>n - k<\/sup><\/p>\r\nwhere p<\/em> denotes the probability of success and q<\/em> = (1 \u2212 p<\/em>) the probability of failure.\r\n

      Note:<\/strong> As a reminder from Chapter 5, Section 5.4, the Number of Combinations of n Objects Taken r at a Time is represented as\u00a0<\/span>nC<\/span>k = $\\frac{n!}{(n-k)!k!}$, where <\/span>n and k are natural numbers. <\/em><\/span><\/span><\/p>\r\n\r\n<\/div>\r\n \r\n\r\nWe use the above formula to solve the following examples.\r\n\r\n \r\n

      \r\n

      Example 7.1.2<\/p>\r\n\r\n<\/header>\r\n

      If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?<\/div>\r\n
      Solution<\/strong><\/div>\r\n
      Let S<\/em> denote the probability of obtaining a head, and F<\/em> the probability of obtaining a tail.<\/div>\r\n
      Clearly, n<\/em> = 10, k<\/em> = 3, p<\/em> = 1\/2, and q<\/em> = 1\/2.<\/div>\r\n
      Therefore:<\/div>\r\n
      \r\n

      Therefore:\u00a0<\/em><\/span><\/p>\r\n

      $10C3=\\frac{10!}{(10-3)!3!}=\\frac{10!}{7!3!}=120$<\/em><\/span><\/p>\r\n\r\n<\/div>\r\n

      b<\/em>(10, 3; 1\/2) = 10C3(1\/2)3<\/sup>(1\/2)7<\/sup> = 120(1\/2)3<\/sup>(1\/2)7<\/sup> = .1172<\/div>\r\n<\/div>\r\n \r\n
      \r\n

      Example 7.1.3<\/p>\r\n\r\n<\/header>\r\n

      If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 6 out of 10 free throws in a game?<\/div>\r\n
      Solution <\/strong><\/div>\r\n
      The probability of making a free throw is 3\/4. Therefore, p<\/em> = 3\/4, q<\/em> = 1\/4, n<\/em> = 10, and k<\/em> = 6.<\/div>\r\n
      Therefore:<\/div>\r\n
      b<\/em>(10, 6; 3\/4) = 10C6(3\/4)6<\/sup>(1\/4)4<\/sup> = .1460<\/div>\r\n<\/div>\r\n \r\n
      \r\n

      Example 7.1.4<\/p>\r\n\r\n<\/header>\r\n

      If a medicine cures 80% of the people who take it, what is the probability that of the eight people who take the medicine, 5 will be cured?<\/div>\r\n
      Solution<\/strong><\/div>\r\n
      Here p<\/em> = .80, q<\/em> = .20, n<\/em> = 8, and k<\/em> = 5.<\/div>\r\n
      b<\/em>(8, 5; .80) = 8C5(.80)5<\/sup>(.20)3<\/sup> = .1468<\/div>\r\n<\/div>\r\n \r\n
      \r\n

      Example 7.1.5<\/p>\r\n\r\n<\/header>\r\n

      If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?<\/div>\r\n
      Solution<\/strong><\/div>\r\n
      If S<\/em> denotes the probability that the chip is defective, and F<\/em> the probability that the chip is not defective, then p<\/em> = .04, q<\/em> = .96, n<\/em> = 60, and k<\/em> = 3.<\/div>\r\n
      b<\/em>(60, 3; .04) = 60C3(.04)3<\/sup>(.96)57<\/sup> = .2138<\/div>\r\n<\/div>\r\n \r\n
      \r\n

      Example 7.1.6<\/p>\r\n\r\n<\/header>\r\n

      If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?<\/div>\r\n
      Solution<\/strong><\/div>\r\n
      If S<\/em> denoted the probability that a person will buy the product, and F<\/em> the probability that the person will not buy the product, then p<\/em> = .15, q<\/em> = .85, n<\/em> = 12, and k<\/em> = 2.<\/div>\r\n
      b(12, 2; .15) = 12C2(.15)2<\/sup>(.85)10<\/sup> = .2924.<\/div>\r\n<\/div>\r\n \r\n

      Practice questions<\/h1>\r\n1.<\/strong> What is the probability of getting three ones if a die is rolled five times?\r\n\r\n2.<\/strong> A basketball player has an 80% chance of sinking a basket on a free throw. In five free throws, what is the probability that he will sink:\r\n

      a. <\/strong>Only one basket?<\/p>\r\n

      a.\u00a0<\/strong>Three baskets?<\/p>\r\n

      c. <\/strong>At least three baskets?<\/p>\r\n3. <\/strong>If a medicine cures 75% of the people who take it, what is the probability that of 30 people who take the medicine:\r\n

      a. <\/strong>25 will be cured?<\/p>\r\n

      b. <\/strong>26 will be cured?<\/p>\r\n

      c. <\/strong>27 will be cured?<\/p>\r\n

      d. <\/strong>At least 25 will be cured?<\/p>\r\n4.<\/strong> The Canadian Food Inspection Agency (CFIA) has found that 5% of the imported spices into Canada are contaminated with pathogenic food-borne bacteria. What is the probability that a batch of 25 imported spices will have:\r\n

      a. <\/strong>One contaminated product?<\/p>\r\n

      b. <\/strong>Two contaminated products?<\/p>\r\n5.\u00a0<\/strong>An executive has determined that for a door-to-door donation initiative, 20% of the households visited will provide a donation. If 10 households are visited, what is the probability that at most 2 will provide a donation?\r\n\r\n ","rendered":"

      <\/div>\n

      Binomial Probability<\/h1>\n

      In this section, we will consider types of problems that involve a sequence of trials, where each trial has only two outcomes, a success or a failure. These trials are independent. That is, the outcome of one does not affect the outcome of any other trial. Furthermore, the probability of success, p<\/em>, and the probability of failure, (1 \u2212 p<\/em>), remains the same throughout the experiment. These problems are called binomial probability<\/strong> problems. Since these problems were researched by a Swiss mathematician named Jacques Bernoulli around 1700, they are also referred to as Bernoulli trials<\/strong>.<\/p>\n

      We give the following definition:<\/p>\n

      \n

      Binomial Experiment<\/strong>: A binomial experiment satisfies the following four conditions:<\/p>\n

        \n
      1. There are only two outcomes, a success or a failure, for each trial.<\/li>\n
      2. The same experiment is repeated several times.<\/li>\n
      3. The trials are independent; that is, the outcome of a particular trial does not affect the outcome of any other trial.<\/li>\n
      4. The probability of success remains the same for every trial.<\/li>\n<\/ol>\n<\/div>\n

         <\/p>\n

        The probability model that we are about to investigate will give us the tools to solve many real-life problems like the ones given below.<\/p>\n

          \n
        1. If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?<\/li>\n
        2. If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 7 out of 10 free throws in a game?<\/li>\n
        3. If a medicine cures 80% of the people who take it, what is the probability that among the 10 people who take the medicine, 6 will be cured?<\/li>\n
        4. If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?<\/li>\n
        5. If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?<\/li>\n<\/ol>\n

          We now consider the following example to develop a formula for finding the probability of k<\/em> successes in n<\/em> Bernoulli trials.<\/p>\n

           <\/p>\n

          \n
          \n

          Example 7.1.1<\/p>\n<\/header>\n

          A baseball player has a batting average of 0.300. If he bats four times in a game, find the probability that he will have:<\/div>\n
          a.<\/strong> four hits<\/div>\n
          b.<\/strong> three hits<\/div>\n
          c.<\/strong> two hits<\/div>\n
          d.<\/strong> one hit<\/div>\n
          e.<\/strong> no hits.<\/div>\n
          Solution <\/strong><\/div>\n
          Let us suppose S<\/em> denotes that the player gets a hit, and F<\/em> denotes that he does not get a hit. This is a binomial experiment because it meets all four conditions. First, there are only two outcomes, S<\/em> or F<\/em>. Clearly the experiment is repeated four times. Lastly, if we assume that the player’s skillfulness to get a hit does not change each time he comes to bat, the trials are independent with a probability of 0.3 of getting a hit during each trial. We draw a tree diagram to show all situations.<\/div>\n
          \n
          \n

          \"Rendered<\/p>\n<\/div>\n<\/div>\n

          Let us first find the probability of getting, for example, two hits. We will have to consider the six possibilities, SSFF , SFSF , SFFS , FSSF , FSFS , FFSS , as shown in the above tree diagram. We list the probabilities of each below.<\/div>\n
          P<\/em>(SSFF) = (0.3)(0.3)(0.7)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\n
          P<\/em>(SFSF) = (0.3)(0.7)(0.3)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\n
          P<\/em>(SFFS) = (0.3)(0.7)(0.7)(0.3) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\n
          P<\/em>(FSSF) = (0.7)(0.3)(0.3)(0.7) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\n
          P<\/em>(FSFS) = (0.7)(0.3)(0.7)(0.3) = (0.3)2<\/sup>(0.7)2<\/sup><\/div>\n
          P<\/em>(FFSS) = (0.7)(0.7)(0.3)(0.3) = (0.3)2<\/sup>(0.7)2 <\/sup><\/div>\n
          Since the probability of each of these six outcomes is (0.3)2<\/sup>(0.7)2<\/sup>, the probability of obtaining two successes is 6(0.3)2<\/sup>(0.7)2<\/sup>.<\/div>\n
          The probability of getting one hit can be obtained in the same way. Since each permutation has one S<\/em> and three F<\/em>‘s, there are four such outcomes: SFFF , FSFF , FFSF , and FFFS.<\/div>\n
          And since the probability of each of the four outcomes is (0.3)(0.7)3<\/sup>, the probability of getting one hit is 4(0.3)(0.7)3<\/sup>.<\/div>\n
          The table below lists the probabilities for all cases, and shows a comparison with the binomial expansion of fourth degree. Again, p<\/em> denotes the probability of success, and q<\/em> = (1 \u2212 p<\/em>) the probability of failure.<\/div>\n
          \n\n\n\n\n
          Outcome<\/td>\nFour Hits<\/td>\nThree hits<\/td>\nTwo Hits<\/td>\nOne hits<\/td>\nNo Hits<\/td>\n<\/tr>\n
          Probability<\/td>\n(0.3)4<\/sup><\/td>\n4(0.3)3<\/sup>(0.7)<\/td>\n6(0.3)2<\/sup>(0.7)2<\/sup><\/td>\n4(0.3)(0.7)3<\/sup><\/td>\n(0.7)4<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

           <\/p>\n

          This gives us the following theorem:<\/p>\n

          \n

          Binomial Probability Theorem<\/strong>:<\/p>\n

          The probability of obtaining k<\/em> successes in n<\/em> independent Bernoulli trials is given by:<\/p>\n

          P<\/em>(n<\/em>, k<\/em>; p<\/em>) = n<\/em>Ckp<\/em>k<\/sup>q<\/em>n – k<\/sup><\/p>\n

          where p<\/em> denotes the probability of success and q<\/em> = (1 \u2212 p<\/em>) the probability of failure.<\/p>\n

          Note:<\/strong> As a reminder from Chapter 5, Section 5.4, the Number of Combinations of n Objects Taken r at a Time is represented as\u00a0<\/span>nC<\/span>k = \"\frac{n!}{(n-k)!k!}\", where <\/span>n and k are natural numbers. <\/em><\/span><\/span><\/p>\n<\/div>\n

           <\/p>\n

          We use the above formula to solve the following examples.<\/p>\n

           <\/p>\n

          \n
          \n

          Example 7.1.2<\/p>\n<\/header>\n

          If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?<\/div>\n
          Solution<\/strong><\/div>\n
          Let S<\/em> denote the probability of obtaining a head, and F<\/em> the probability of obtaining a tail.<\/div>\n
          Clearly, n<\/em> = 10, k<\/em> = 3, p<\/em> = 1\/2, and q<\/em> = 1\/2.<\/div>\n
          Therefore:<\/div>\n
          \n

          Therefore:\u00a0<\/em><\/span><\/p>\n

          \"10C3=\frac{10!}{(10-3)!3!}=\frac{10!}{7!3!}=120\"<\/em><\/span><\/p>\n<\/div>\n

          b<\/em>(10, 3; 1\/2) = 10C3(1\/2)3<\/sup>(1\/2)7<\/sup> = 120(1\/2)3<\/sup>(1\/2)7<\/sup> = .1172<\/div>\n<\/div>\n

           <\/p>\n

          \n
          \n

          Example 7.1.3<\/p>\n<\/header>\n

          If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 6 out of 10 free throws in a game?<\/div>\n
          Solution <\/strong><\/div>\n
          The probability of making a free throw is 3\/4. Therefore, p<\/em> = 3\/4, q<\/em> = 1\/4, n<\/em> = 10, and k<\/em> = 6.<\/div>\n
          Therefore:<\/div>\n
          b<\/em>(10, 6; 3\/4) = 10C6(3\/4)6<\/sup>(1\/4)4<\/sup> = .1460<\/div>\n<\/div>\n

           <\/p>\n

          \n
          \n

          Example 7.1.4<\/p>\n<\/header>\n

          If a medicine cures 80% of the people who take it, what is the probability that of the eight people who take the medicine, 5 will be cured?<\/div>\n
          Solution<\/strong><\/div>\n
          Here p<\/em> = .80, q<\/em> = .20, n<\/em> = 8, and k<\/em> = 5.<\/div>\n
          b<\/em>(8, 5; .80) = 8C5(.80)5<\/sup>(.20)3<\/sup> = .1468<\/div>\n<\/div>\n

           <\/p>\n

          \n
          \n

          Example 7.1.5<\/p>\n<\/header>\n

          If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?<\/div>\n
          Solution<\/strong><\/div>\n
          If S<\/em> denotes the probability that the chip is defective, and F<\/em> the probability that the chip is not defective, then p<\/em> = .04, q<\/em> = .96, n<\/em> = 60, and k<\/em> = 3.<\/div>\n
          b<\/em>(60, 3; .04) = 60C3(.04)3<\/sup>(.96)57<\/sup> = .2138<\/div>\n<\/div>\n

           <\/p>\n

          \n
          \n

          Example 7.1.6<\/p>\n<\/header>\n

          If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?<\/div>\n
          Solution<\/strong><\/div>\n
          If S<\/em> denoted the probability that a person will buy the product, and F<\/em> the probability that the person will not buy the product, then p<\/em> = .15, q<\/em> = .85, n<\/em> = 12, and k<\/em> = 2.<\/div>\n
          b(12, 2; .15) = 12C2(.15)2<\/sup>(.85)10<\/sup> = .2924.<\/div>\n<\/div>\n

           <\/p>\n

          Practice questions<\/h1>\n

          1.<\/strong> What is the probability of getting three ones if a die is rolled five times?<\/p>\n

          2.<\/strong> A basketball player has an 80% chance of sinking a basket on a free throw. In five free throws, what is the probability that he will sink:<\/p>\n

          a. <\/strong>Only one basket?<\/p>\n

          a.\u00a0<\/strong>Three baskets?<\/p>\n

          c. <\/strong>At least three baskets?<\/p>\n

          3. <\/strong>If a medicine cures 75% of the people who take it, what is the probability that of 30 people who take the medicine:<\/p>\n

          a. <\/strong>25 will be cured?<\/p>\n

          b. <\/strong>26 will be cured?<\/p>\n

          c. <\/strong>27 will be cured?<\/p>\n

          d. <\/strong>At least 25 will be cured?<\/p>\n

          4.<\/strong> The Canadian Food Inspection Agency (CFIA) has found that 5% of the imported spices into Canada are contaminated with pathogenic food-borne bacteria. What is the probability that a batch of 25 imported spices will have:<\/p>\n

          a. <\/strong>One contaminated product?<\/p>\n

          b. <\/strong>Two contaminated products?<\/p>\n

          5.\u00a0<\/strong>An executive has determined that for a door-to-door donation initiative, 20% of the households visited will provide a donation. If 10 households are visited, what is the probability that at most 2 will provide a donation?<\/p>\n

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