probability that 3rd head occurred on 9th trial If it is known that in a series of coin tosses, the 4th head occurred on the 12th trial, what is the probability that the 3rd head occurred on the 9th trial?
Here's what I did, but I don't know if it's right:
$$\text{total outcomes with $12^\text{th}$ trial being $4^\text{th}$ head} = {11 \choose 3} = 165$$
$$\text{total outcomes with $9^\text{th}$ trial being $3^{\text{rd}}$ head} = {8 \choose 2}= 28$$
$$P(\text{$3^\text{rd}$ head on $9^\text{th}$ trial $\mid$ $4^\text{th}$ head on $12^\text{th}$ trial}) = \frac{28}{165}$$
 A: You answer is slightly incorrect. 
You need to take into account that the 10th, 11th and 12th trial are T, T, H accordingly. Since this can only occur in 1 way, we'd multiply 28 by 1 to still get 28.
A: Let us use the ordinary conditional probability formalism. Let $A$ be the event the fourth head occurred on the $12$th trial, and let $B$ be the event the third head occurred on the $9$-th trial. We want $\Pr(B|A)$. We have as usual
$$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$
Now compute the probabilities on the right. 
The $4$-th head occurs on the $12$th trial if there are $3$ heads in the first $11$, and then a head. The probability of this is $\dbinom{11}{3}(1/2)^{11}(1/2)$.
Now we find $\Pr(A\cap B)$. The probability of $B$ is $\dbinom{8}{2}(1/2)^9$. Given that $B$ happened, the probability of $A$ is the probability that in coin tossing, the first head occurs on the third trial. This is just the probability of TTH, which is $(1/2)^3$. Thus $\Pr(A\cap B)=\dbinom{8}{2}(1/2)^{12}$.
Remark: When we do the division, the powers of $1/2$ cancel, and we get precisely your answer. That answer was obtained in a combinatorially nicer way, but there is a certain amount of security in using standard conditional probability machinery.
