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(a) A fair coin is tossed until the first H occurs. Compute the probability that three tosses are required.

(b) A fair coin is tossed until the second H occurs. Compute the probability that five tosses are required.

(c) A coin with $P(\text{heads})=\frac{2}{3}$ is tossed until the third T. Find the probability of five heads.

for (a) there is only possible case : TTH , Since $3$ Tosses are required, The probability is $1\times (0.5)\times (0.5)\times (0.5)$

for (b) there are $4$ possibilities then the probability will be $4\times (0.5)^5$

for (c) I have no clue what to do with the third case.

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  • $\begingroup$ To avoid downvotes and closing add your own effort to the question and tell us where you eventually got stuck. $\endgroup$ – drhab Sep 24 '18 at 6:47
  • $\begingroup$ for (a) there is only possible case : TTH , Since 3 Tosses are required, The probability is 1*(0.5)*(0.5)*(0.5) for (b) there are 4 Possibilities then the probability will be 4*(0.5)^5 for (c) I have no clue what to do with the third case. $\endgroup$ – dark lion Sep 24 '18 at 6:56
  • $\begingroup$ @darklion My 2 cents... for c) check out the probability mass function of a negative binomial random variable. You actually used it already for a) and b). $\endgroup$ – HJ_beginner Sep 24 '18 at 7:16
  • $\begingroup$ @darklion please edit your post to include your working :) $\endgroup$ – user107224 Sep 24 '18 at 7:53
  • $\begingroup$ You may want to check the negative binomial distribution $\endgroup$ – J. R. C. Sep 24 '18 at 9:43
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Hints for (c)

  • To get the probability of five Hs by the third T, you need five of the first seven tosses to be Hs and two to be Ts, while the eighth toss must be T.

  • The distribution of the first seven tosses is binomial

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    $\begingroup$ I got it . answer is 21*(1/3)^3*(2/3)^5 $\endgroup$ – dark lion Sep 24 '18 at 8:59
  • $\begingroup$ @darklion - indeed, just over $0.1$ $\endgroup$ – Henry Sep 24 '18 at 9:22

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