Stuck on this question.

The experiment of tossing a fair coin until three consecutive heads appear is performed. Let X be the number of tosses, and Y be the number of tails that appear. Find the probability p(Y = 1).

I tried listing out the possibilities where Y=1:


So I thought maybe it's 3/(2^6) but that's not listed as an answer.


1 Answer 1


If the first toss is a T, then the event $Y=1$ happens if the first $4$ tosses are THHH. This has probability $\frac{1}{16}$.

If the first toss is a H, then $Y=1$ can happen in $2$ ways: (i) the second toss is a T and then we get HHH or (ii) the second toss is a H, the next is a T, and then we get HHH. Note that (i) has probability $\frac{1}{32}$ and (ii) has probability $\frac{1}{64}$.

Add up our three probabilities, and simplify.

Remark: Your idea was right, there was just a little error in computing the probabilities.

  • $\begingroup$ Seems to make sense to me. Doesn't match any of the answers given, however. (1/16 + 1/32 + 1/64 = 7/64). Answer choices are 7/18, 3/32, 25/32, 5/6, 2/3. Wrong sample question? $\endgroup$ Mar 19, 2015 at 23:07
  • $\begingroup$ The unnecessary definition of the random variable $X$ makes me wonder whether what you have is the full question. But the probability that $Y=1$ is clear. If while solving we forgot about HHTHHH, we would get $3/32$, one of the given answers. But of course that would be wrong. $\endgroup$ Mar 20, 2015 at 5:06
  • $\begingroup$ X comes into play in subsequent questions. Don't we have to consider cases like TTTTTTTTTTHHH in the denominator? $\endgroup$ Mar 21, 2015 at 4:13
  • $\begingroup$ @SeanBollin: Yes, the interpretation as a conditional probability problem is fully reasonable. However, the probability that sometime or other there will be $3$ consecutive heads is $1$, so the answer is $7/64$. $\endgroup$ Mar 21, 2015 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.