91Ó°ÊÓ

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 .

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

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

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

$=0.132$

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.

$Yâ‰\yen 5$

$P(Yâ‰\yen 5)=\underset{1=5}{\overset{3}{∑}}P(Y=k)$

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

$=$$0.7414$

The final answer is$P\left(\left(Yâ‰\yen 5\right)\right)=$ $0.7414$

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$0.1561$