Conditional probability/Bayes Theorem Question I'm studying for my stat midterm and I got stuck on this problem. Could anyone give me some feedback?
A new drug for leukemia works 25% of the time in patients 55 and older, and 50% of the time in patients younger than 55. A test group has 17 patients 55 and older and 12 patients younger than 55.
A subgroup of 4 patients are chosen and the drug is administered to each. What is the probability that the drug works in all of them?
The answer is supposedly 1.52%.
I made a tree diagram, W = drug works, 55+ = over 55, 55- = under 55, and determined that
P(W) = P(W|55+)P(55+) + P(W|55-)P(55-) = .353
I then took .353^4 to get 1.553% but apparently this answer is wrong. How do I get the correct answer?

We can do part of a slightly tedious calculation. We draw a random sample of $4$ people from the $17$ over $55$'s and from the $12$ under $55$'s, that is, from $29$ people.  Recall that the probability that the drug works for an over $55$ is $a=0.25$, and works for an under $55$ with probability $b=0.50$.
The probability of having everybody be over $55$ is $\frac{\binom{17}{4}\binom{12}{0}}{\binom{29}{4}}$. The probability the drug works for all of them is $a^4$. So that branch of the tree has probability $\frac{\binom{17}{4}\binom{12}{0}}{\binom{29}{4}}a^4$. 
The probability that there are $3$ over $55$ and $1$ under is $\frac{\binom{17}{4}\binom{12}{0}}{\binom{29}{4}}$.  The probability the drug works on all of them is $a^3b$. So that branch of the tree has probability $\frac{\binom{17}{4}\binom{12}{0}}{\binom{29}{4}}a^3b$.
There are three other terms that you need to write down. Now add up the five probabilities obtained. 
