Questionnaire probability When sent a questionnaire, the probability is .5 that any particular individual to whom it is sent will respond immediately to that questionnaire. For an individual who did not respond immediately, there is a probability of .4 that the individual will respond when sent a follow-up letter. If the questionnaire is sent to 4 persons and a follow-up letter is sent to any of the 4 who do not respond immediately, what is the probability that at least 3 never respond?
The probability that at least 3 never respond is 
P[3 never respond]+P[4 never respond]
P[4 never respond] is just $(.3)^4$. To find P[3 never respond], it's P[1 doesn't respond from first round] + P[1 doesn't respond from the second round]. 
So my steps:
P[3 never respond]+P[4 never respond]
[1 doesn't respond from first round] + P[1 doesn't respond from the second round] +P[4 never respond]
$(.5)(.5)^3(.6)^3$+$(.5^4)(.6)^3(0.4)$+$(.3)^4$
=$(.5)^4(.6)^3$+$(.5^4)(.6)^3(0.4)$+$(.3)^4$
However, according to the book, the answer is 
4[$(.5)^4(.6)^3$]+4[$(.5^4)(.6)^3(0.4)$]+$(.3)^4$
=$4(.3)^3(.7)$+$(.3)^4$
Where did I go wrong? Where did the 4's come from? How do you get from the second to last step to the last one?
 A: Let $X$ be the number of those who respond immediately and Let $Y$ be the number of those who never respond. The probability that at least $3$ will never respond is
$$P(Y\ge 3)=P(Y=3)+P(Y=4).$$
So far, so good. Then
$$P(Y=3)=P(Y=3\mid X=0)P(X=0)+P(Y=3 \mid X=1)P(X=1).$$
and
$$P(Y=4)=P(Y=4\mid X=0)P(X=0).$$
Now


*

*$P(X=0)=\frac{1}{2^4}=\frac{1}{16},$

*$P(X=1)={4 \choose 1}\frac{1}{2^4}=\frac{1}{4},$


and 


*

*$P(Y=3\mid X=0)={4 \choose 3}\left(\frac{6}{10}\right)^3\frac{4}{10}=4\left(\frac{6}{10}\right)^3\frac{4}{10},$

*$P(Y=3\mid X=1)=\left(\frac{6}{10}\right)^3,$

*$P(Y=4\mid X=0)=\left(\frac{6}{10}\right)^4.$


So,
$$P(Y\ge 3)=4\left(\frac{6}{10}\right)^3\frac{4}{10}\frac{1}{16}+\left(\frac{6}{10}\right)^3\frac{1}{4}+\left(\frac{6}{10}\right)^4\frac{1}{16}=0.0837.$$
A: I'm many years late, but this is for anyone searching for the solution in today's date since this question gave me a hard time whilst the solution turned out to be fairly easy. Just requires a little logic.
We need AT LEAST 3 people who never respond. That means the event of 4 people who never respond with the addition of P(3 people never respond, 1 responds immediately) and P(3 people who never respond, 1 responds upon follow-up).
We all understand the first part.
0.4 is the probability of people who respond upon sending a follow-up, which means 0.6 is the probability of people who don't respond after sending a follow-up because we subtract 0.4 from 1.
Then, 0.5*0.6 = 0.3 becomes the probability of people who never respond.
Raising this to the power of four solves the first part, where four people never respond.
Now, the second part.
There are four people out of whom 1 responds immediately or upon follow-up. So, that's 4C1 people, which equals 4. That's where the four comes from since this specific question is in the chapter of Combinatorial Principles. Frankly, this also the only part of the question where we use the learning of this topic, rest is just some arithmetic. We will multiply this with 0.7, which comes by subtracting 0.3 from 1. Since 0.3 are the people who never respond, the complement of it are the people who respond (whether immediately or after sending a follow up). Uptil now we have 4*(0.7).
We know that for this case, there are 3 people who never respond. That gives us (0.3)^3.
Combining both of these gives 4*(0.7)(0.3).
Our final answer becomes the addition of both the answers above.
(0.3)^4 + [4(0.7)*(0.3)^3]
