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We have a biased coin such that $P(head) = 0.25$ and $P(Tails) = 0.75$, suppose we flip coin until we observe third heads. What is the probability that we stop flipping after exactly ten flips?.

I thought that let us fix of getting a third head at last that is at 10th flip, so that we would stop there, and the remaining - getting two heads can be accommodated in the 9 trials. so there are $9$ choose 2 ways of getting two heads so the probability that we stop flipping after exactly ten flips is $^9C_{2}$ . $\frac{1}{4}^3$.$\frac{3}{4}^7$. Is this correct?

EDIT - Now the probability of getting exactly 3 heads? I got it to be $^{10} C_{3} \frac{1}{4}^3 \frac{3}{4}^7$. Should we get the same as the previous one? any reason why they should/should not be same?

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    $\begingroup$ I think you switched $P(H),P(T)$ but the approach is good. $\endgroup$ – lulu Oct 1 '18 at 16:13
  • $\begingroup$ oh i see now! thanks! $\endgroup$ – BAYMAX Oct 1 '18 at 16:13
  • $\begingroup$ @lulu please see the edit $\endgroup$ – BAYMAX Oct 1 '18 at 16:30
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    $\begingroup$ Your probability for exactly 3 heads is right as well. It should be obvious why the results have to be different. In the first case the outcome of the last flip is fix and in the second case the outcome of the last flip is not fix. $\endgroup$ – callculus Oct 1 '18 at 16:31
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    $\begingroup$ Edit looks good. The second question differs from the first because you aren't requiring that the last toss be $H$. $\endgroup$ – lulu Oct 1 '18 at 16:32
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$$\sum\limits_{k=3}^{10} {10 \choose k} \left( {1 \over 4} \right)^k \left( {3 \over 4} \right)^{10-k} = 0.474$$

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