Binomial Distribution Probablity

Question

An interviewer is given a list of people she can interview. If the interviewer needs to interview 6 people, and if each person (independently) agrees to be interviewed with probability 3/4, 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) 6 people and (b) 10 people? For part (b), what is the probability that the interviewer will speak to exactly (c) 7 people and (d) 9 people on the list?

For part (b), I thought I would use the Bernoulli Distribution and find out the value when X = 6 (since she wants 6 successes).

I would have P(X=6) and get some answer. However the answer is P(X>=6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)

Why are they getting the sum for all bigger than 6 and not just 6?

EDIT: for part c) I think I'm supposed to use a negative binomial but I'm stuck after that.

Answer 1

a) The interviewer has only 6, so all of them should accept, thus $P(success) = 0.75^6$.

b) Since she needs at least 6 people to interview if the Bernoulli random variable $X$ is at least 6, s/he is good, thus $P(success) = Pr(X\geq6)$.

c) $Pr(X=7) = C(10,7) \cdot 0.75^7 \cdot 0.25^3 $.

d) $Pr(X=9) = C(10,9) \cdot 0.75^9 \cdot 0.25^1 $.

where $C(n,k)$ is the number of different possibilities to choose k objects from n.

Answer 2

Your first question:

They use the sum because she needs at least 6 people to accept, so if she gets 6, 7, 8, 9 or 10, she'll be ok. If all of them accept to be interviewed she'll just select 6 of them. If only 5 of them accept, she doesn't get the 6 interviews needed.

Your second question:

Either you go with the probability that exactly 7 of them accept to be interviewed $P[X=7]$, or you go with zero because she'll only speak to the 6 she needs and not more than that. I find the problem statement for c and d sort of ambiguous because of the word "speak".