A professor has learned that two students in her class of 15 will cheat on the exam. She decides to focus her attention on five randomly chosen students during the exam. a. What is the probability that she finds at least one of the students cheating? (Round your final answers to 4 decimal places.) Probability ? b. What is the probability that she finds at least one of the students cheating if she focuses on seven randomly chosen students? (Round your final answers to 4 decimal places.) Probability ?

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A professor has learned that two students in her class of 15 will cheat on the exam. She decides to focus her attention on five randomly chosen students during the exam.   a. What is the probability that she finds at least one of the students cheating? (Round your final answers to 4 decimal places.) Probability ?   b. What is the probability that she finds at least one of the students cheating if she focuses on seven randomly chosen students? (Round your final answers to 4 decimal places.) Probability ?

Kindlein6h · Accepted Answer

Step 1   Provided information is ;   p=Probability that student will cheat  q=1−p=1−215=1315  Probability density function of Binomial distribution is given by ;  P(X=x)=nCxpxqn−x;x=0,1,2,…n  Step 2   Calculation:   Given: p=215,q=1315  For Part a] :   Given: n=5   Probability that at least one of the students cheating i.e P(x≥1) is given by ;   P(x≥1)=1−P(x&amp;lt;1)   =1−P(x=0)  =1−[5C0(215)0(1315)5−0]  =1−0.4889  =0.5111  Probability that at least one of the students cheating if she focuses on 5 randomly (n=5) chosen students i.e P(x≥1) is equal to 0.5111   For Part b] :    Given n=7   Probability that at least one of the students cheating  i.e P(x≥1) is given by ;   P(x≥1)=1−P(x&amp;lt;1)   =1−P(x=0)  =1−[7C0(215)0(1315)7−0]  =1−0.3672  =0.6328  Probability that at least one of the students cheating if she focuses on 7 randomly (n=7) chosen students i.e P(x≥1) is equal to 0.6328

macalpinee3 · Accepted Answer

Step 1  The probability of the variable X that says the number of students cheating follow a Binomial distribution, because there are:  10 randomly chosen students or n identical and independent events  A probability of 927 that the student is going to cheat or a probability of 13 of success  So, the probability P(x) that x students are going to cheat is calculated as:  P(x)=n!x!(n−x)!×px×(1−p)n−x  P(x)=10!x!(10−x)!×13x×(1−13)10−x  Then, the probability P that she finds at least one of the students cheating is calculated as:  P=P(x≥1)=P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)  That also can be calculated as:  P=P(x≥1)=1−P(x&amp;lt;1)=1−P(0)  Therefore, P(0) is calculated as:  P(0)=10!0!(10−0)!×130×(1−13)10−0  P(0)=0.0173  Finally, P is equal to:  P=1−0.0173=0.9827

A professor has learned that two students in her class of 15

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