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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:

THHH HTHHH HHTHHH

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

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Sean Bollin
    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. $\endgroup$
    – André Nicolas
    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? $\endgroup$
    – Sean Bollin
    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 $7/64$. $\endgroup$
    – André Nicolas
    Mar 21, 2015 at 4:30

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