Suppose we toss a fair coin until we get exactly 2 heads. What is
the probability that exactly 5 tosses are required?
My try:
We have to make sure that the first 4 tosses does not have 2 heads and the last toss must be a head. That is, the first 4 tosses need to contain 1 head and 3 tails. The probability of this event is $\frac{4}{2^4}=1/4$. Then the probability of 5th toss is head is $1/2$. Hence, in the end the answer is $\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$.
Am I correct?
Best Answer
The result is correct, but one of the intermediate steps is incorrect. You first write "the first $4$ tosses [do] not have $2$ heads", and then "That is, the first $4$ tosses need to contain $1$ head and $3$ tails". That's not the same thing; the second formulation is correct, whereas the first formulation would also include results with $0$ heads in the first $4$ tosses.