I have the following question:
An interviewer is given a list of 8 people. Each person agrees to be interviewed with probability $2/3$. What is the probability that she will speak to exactly a) 6 people, b) 7 people?
I tried this problem by saying: Defind $X = $ the number of people who agree to be interviewed out of 8. So: $$X = \mathcal{Binomial}(8,\frac{2}{3})$$
I want for part a:
$$P(X=6) = {^8\mathrm C_6}(2/3)^6(1/3)^2 = 1792/6561$$
and for part b:
$$P(X=7) = {^8\mathrm C_7}(2/3)^7(1/3)^1 = 1024/6561$$
Which according to the book is incorrect. The true answer, for both parts is $160/729$.
Any ideas or clues???
If she interviews exactly 6 people, then 6 will have said yes and 2 will have said no (independently). For each size-6 subset of 8 people, this occurs with probability $(2/3)^6 \cdot (1/3)^2$, and there are $\binom 86 = 28$ such subsets.
And if she interviews exactly 7, then 7 will have said yes and 1 will have said no. For each subset of size 7, this occurs with probability $(2/3)^7 \cdot (1/3)$, and there are $\binom 87 = 8$ such subsets. Thus $$ \mathrm{Prob}(X = 6) = 28 \cdot (2/3)^6 \cdot (1/3)^2 = \dfrac{1792}{6561} \simeq 0.273 $$ and $$ \mathrm{Prob}(X = 7) = 8 \cdot (2/3)^7 \cdot (1/3) = \dfrac{1024}{6561} \simeq 0.156 $$