Coin toss problem, get exactly 2 heads in 5 tosses

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?

• Is your question what is the probability that the third head is obtained on the fifth toss? – N. F. Taussig Sep 3 '15 at 12:12
• @N.F.Taussig Sorry a typo. Corrected! – spatially Sep 3 '15 at 12:13

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.

This really sounds like you are saying the following:

" The probability of the number X of Bernoulli trials needed to get n successes "

If that is the case we can just apply the negative binomial formula:

$$C^{n + r -1}_{r-1} p^{r} (1-p)^{n}$$

$r$ is the number of successes and $n$ is the number of failures so $r=2, n=3$ which yields:

• $n+r-1= 2+3 -1=4$
• $C^{n+r-1}_{r-1} = C^{4}_{1} = 4=6$

$$C^{4}_{1} (0.5)^{2} (0.5)^{3}$$

$$4(0.5)^{2} (0.5)^{3} = \frac{4}{2^{5}} = \frac{1}{8}$$

• @Jerryoverone: No, this answer is wrong. The number of successes is $r=3$, the number of failures is $n=2$; if you plug in those numbers, you get $1/8$ as in your correct solution. More generally, I wouldn't trust an answer that derives a result contradicting yours but doesn't explain the discrepancy. – joriki Sep 3 '15 at 12:45
• your comment above does not clarify where the answer goes wrong and I disagree the number of succeses is 2. Explicitly speaking the number of succeses is 2 which is the number of times it lands head and the number of failures is 3 being the number of times it does not land head. The mistake was in in $C^{n+r-1}_{r}$ corrected to $C^{n+r-1}_{r-1}$ reference. – Francisco Vargas Sep 3 '15 at 13:15