Probability of winning chocolates in throwing game In urn there are 12 balls from which 7 are green.John draws at random 4 balls.Then he hits a target with the green balls he drew(if there are any).His probability of hitting successfully is $\frac{1}{4}$ in every shot and if he succeeds he wins a chocolate for every successful hit.
a)What is the probability John to win exactly 2 chocolates?
b)if a) happened,what is the probability he drew exactly 3 green balls from the urn?
How I solved it I think : Let $A$ be the event John wins 2 chocolates, and let $H_{i}$ for $i$ in ${[0,1,2,3,4]}$ be the event - John drew exactly $i$ green balls.
So for a) we seek $P(A|H_{1} \cup H_{2} \cup H_{3} \cup H_{4}) =P(A|H_{0})P(H_{0})+(A|H_{1})P(H_{1})+P(A|H_{2})P(H_{2})+P(A|H_{3})P(H_{3})+P(A|H_{4})P(H_{4})$.
The cases where $i=0$ and $i=1$ , the probability  is $0$ so we need to find 
$P(A|H_{2})P(H_{2})+P(A|H_{3})P(H_{3})+P(A|H_{4})P(H_{4})$.
$P(H_{2}) = \frac{ \binom{7}{2}\binom{5}{2}}{\binom{12}{4}}$ ,
$P(H_{3}) = \frac{ \binom{7}{3}\binom{5}{1}}{\binom{12}{4}}$ ,$P(H_{4}) = \frac{ \binom{7}{4}\binom{5}{0}}{\binom{12}{4}}$ ,
$P(A|H_{2})=\frac{1}{4}\frac{1}{4}$ - since we want him to succeed two times
$P(A|H_{3})=3(\frac{1}{4}\frac{1}{4}\frac{3}{4})$  - since we want two success and 1 failure multiplied by 3 for the permutations
$P(A|H_{4})=6(\frac{1}{4}\frac{1}{4}\frac{3}{4}\frac{3}{4})$ 
Now we just plug them in above and find the answer
for b) I think we seek  $P(H_{3}|A)=?$ which can be found applying Bayes's theorem and plugging in the values from a)
Is my solution correct?

Note that John can only win 2 chocolates if he drew at least 2 green balls from the urn. So the first step is to find the probs of drawing 2, 3 and 4 greens from the urn.
We have $$p(2G)= 6\frac{7}{12}\frac{6}{11}\frac{5}{10}\frac{4}{9}=\frac{42}{99}$$ Similarly $$p(3G)=\frac{35}{99},p(4G)=\frac{7}{99}$$
We also need to work out $$p(2c|2G)=\frac{1}{4}\frac{1}{4}=\frac{8}{128},\quad p(2c|3G)=3\frac{1}{4}\frac{1}{4}\frac{3}{4}=\frac{18}{128},\quad p(2c|4G)=6\frac{9}{256}=\frac{3}{128}$$
So now we can apply Bayes to get that $$p(3G|2c)=\frac{p(2c|3G)p(3G)}{p(2c|2G)p(2G)+p(2c|3G)p(3G)+p(2c|4G)p(4G)}$$ $$=\frac{18\cdot35}{8\cdot42+18\cdot35+3\cdot7}=\frac{630}{987}\approx0.64$$
