Question

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?

Answer 1

Answer:

a) 0.1317

b) 0.7413

c) 0.2731

d) 0.1561

Step-by-step explanation:

Let's start defining the random variable.

X : ''The number of people that she interviews''

X ~ Bi (n,p)

This means, X can be modeled as a Binomial random variable where p is the success probability and n is the number of independent Bernoulli experiments that take place. In this case, n is the number of interviews that she makes.

The probability function for X is :

$P(X=x)=(nCx).p^{x}(1-p)^{n-x}$

Where nCx is the combinatorial number defines as :

$nCx=\frac{n!}{x!(n-x)!}$

For a)

We are looking for the probability of she interviews 5 people over 5 she interviews ⇒

$P(X=5)=(5C5)(\frac{2}{3})^{5}(\frac{1}{3})^{0}=\frac{32}{243}=0.1317$

b)  In this part , we are looking for P(X ≥ 5) when n = 8 (Number of people she interviews)

$P(X\geq 5)= P(X=5)+P(X=6)+P(X=7)+P(X=8)$

$P(X\geq 5)=(8C5)(\frac{2}{3})^{5}(\frac{1}{3})^{3}+(8C6)(\frac{2}{3})^{6}(\frac{1}{3})^{2}+(8C7)(\frac{2}{3})^{7}(\frac{1}{3})^{1}+(8C8)(\frac{2}{3})^{8}(\frac{1}{3})^{0}=0.7413$

c)  Interview 6 people of 8 :

$P(X=6)=(8C6)(\frac{2}{3})^{6}(\frac{1}{3})^{2}=0.2731$

d) Interview 7 people of 8 :

$P(X=7)=(8C7)(\frac{2}{3})^{7}(\frac{1}{3})^{1}=0.1561$