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?

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,
$$P(Y\ge 3)=4\left(\frac{6}{10}\right)^3\frac{4}{10}\frac{1}{16}+\left(\frac{6}{10}\right)^3\frac{1}{4}+\left(\frac{6}{10}\right)^4\frac{1}{16}=0.0837.$$
