# 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:

$$\text{total outcomes with 12^\text{th} trial being 4^\text{th} head} = {11 \choose 3} = 165$$

$$\text{total outcomes with 9^\text{th} trial being 3^{\text{rd}} head} = {8 \choose 2}= 28$$

$$P(\text{3^\text{rd} head on 9^\text{th} trial \mid 4^\text{th} head on 12^\text{th} trial}) = \frac{28}{165}$$

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There were 2 heads in the first 8 trials; the 9th trial was a head; the 10th and 11th trails were tails. Analyzing it that way, can you work out the probability? – Gerry Myerson Jan 23 '13 at 5:39

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.

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