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

(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.

• To avoid downvotes and closing add your own effort to the question and tell us where you eventually got stuck. – drhab Sep 24 '18 at 6:47
• 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. – dark lion Sep 24 '18 at 6:56
• @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). – HJ_beginner Sep 24 '18 at 7:16
• @darklion please edit your post to include your working :) – user107224 Sep 24 '18 at 7:53
• You may want to check the negative binomial distribution – J. R. C. Sep 24 '18 at 9:43

• @darklion - indeed, just over $0.1$ – Henry Sep 24 '18 at 9:22