An interviewer is given a list of people she can interview. If the interviewer needs to interview 5 people, and if each person (independently) agrees to be interviewed with probability 2 3 , what is the probability that her list of people will enable her to obtain her necessary number of interviews if the list consists of

(a) 5 people and

(b) 8 people? For part (b), what is the probability that the interviewer will speak to exactly

(c) 6 people and

(d) 7 people on the list?

If the interviewer needs to interview 5 people,and if each person agrees to be interviewed with probability of 2/3.

consider X is the random variable that represents the number of people that agree to be interviewed .

$P(X=5)=\left(\begin{array}{l}n\\ 5\end{array}\right)p^5(1-p{)}^{n-5}$

$=\left(\begin{array}{l}5\\ 5\end{array}\right){\left(\frac{2}{3}\right)}^5{\left(1-\frac{2}{3}\right)}^{5-5}$

$=1\times {\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^0$

$=0.132$

The final answer is$P\left(X=5\right)$$0.132$

Consider Y is the random variable that represents the number of people that agree to be interviewed .

The number of people is 8.

The probability of success is 2/3.

$Y\ge 5$

$P(Y\ge 5)=\sum _{1=5}^{3}P(Y=k)$

$=\left[\begin{array}{l}P(Y=5)+P(Y=6)+\\ P(Y=7)+P(Y=8)\end{array}\right]$

$=$$0.7414$

The final answer is$P\left(\left(Y\ge 5\right)\right)=$ $0.7414$

Consider Y is the random variable that represents the number of people . the list consists of 6 people. the probability of success is 2/3

$P(Y=6)=\left(\begin{array}{l}n\\ 6\end{array}\right)p^6(1-p{)}^{n-6}$

$=\left(\begin{array}{l}8\\ 6\end{array}\right){\left(\frac{2}{3}\right)}^6{\left(1-\frac{2}{3}\right)}^{8-6}$

$=0.2731$

The final answer is$P\left(Y=6\right)=$$0.2731$

Consider Y is the random variable that represents the number of people . the list consists of 7 people. the probability of success is 2/3

$=\left(\begin{array}{l}8\\ 7\end{array}\right){\left(\frac{2}{3}\right)}^7{\left(1-\frac{2}{3}\right)}^{8-7}$

$=0.1561$

The final answer is$0.1561$