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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?

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

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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:

\begin{equation} C^{n + r -1}_{r-1} p^{r} (1-p)^{n} \end{equation}

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

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

\begin{equation} C^{4}_{1} (0.5)^{2} (0.5)^{3} \end{equation}

\begin{equation} 4(0.5)^{2} (0.5)^{3} = \frac{4}{2^{5}} = \frac{1}{8} \end{equation}

Which agrees with your answers sorry for the typos :)

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  • $\begingroup$ So my answer is wrong? $\endgroup$ – spatially Sep 3 '15 at 12:42
  • $\begingroup$ @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. $\endgroup$ – joriki Sep 3 '15 at 12:45
  • $\begingroup$ The answer is still wrong after the edit; please see my comment above. $\endgroup$ – joriki Sep 3 '15 at 13:07
  • $\begingroup$ 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. $\endgroup$ – Francisco Vargas Sep 3 '15 at 13:15
  • $\begingroup$ It depends on what you call a success and a failure. Your previous formula would have been correct with successes and failures swapped, but changing the formula also solved the problem. $\endgroup$ – joriki Sep 3 '15 at 14:50

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