Question

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention on four randomly chosen students during the exam.
a. What is the probability that he finds at least one of the students cheating? Round your intermediate calculations and final answers to 4 decimal places.Probability_____b. What is the probability that he finds at least one of the students cheating if she focuses on four randomly chosen students? Round your intermediate calculations and final answers to 4 decimal places.Probability_____

Answer 1

Answer:

a

$P(X \ge 1) = 0.509$

b

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      $P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      $P(X \ge 1) = 1 - P(X \le 0)$

Here  

   $P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

   $P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

   $P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

    $P(X \le 0) = 0.491$

Hence

    $P(X \ge 1) = 1 - 0.491$

     $P(X \ge 1) = 0.509$

Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

    $P(X \ge 1) = 1 - P(X \le 0)$

   $P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$

Here n =  6

So

    $P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]$

    $P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]$

    $P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]$

    $P(X \ge 1) =1- 0.3193$

    $P(X \ge 1) = 0.6807$