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A fair coin is tossed $20$ times. Find the probability that exactly $6$ heads were tossed one of which was tossed on the first toss and one of which was tossed on the last toss, and that no consecutive tosses were heads.

Typically, the probability would be ${20 \choose 6}(1/2)^{6}(1/2)^{14}$.

However, we are restricted in that we cannot include all combinations of successes and failures.

Would the adjusted answer be:

${10 \choose 4}(0.5)^4 (1.5)^{14}$?

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Just count how many of the $2^{20}$ possible sequences satisfy the conditions. Every legal sequence must have the form HT...H, such that the "..." consists of 4 "HT" combinations and 9 "T"s that are not part of such a combination.

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    @lord12, no. I assume your idea is to place all the T's there an put a H next to$4$of them. Which is fine as far as it goes, but you're counting the possible position for H's wrong. Both sequences starting with HTH... and sequences ending with ...HTH. So, starting with HT$^{14}$H, there are 13 (not 12) places you can put a H, and choosing 4 of them gives $\binom{13}{4}$ possibilities.2011-09-27