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Consider a coin for which the $P(\text{heads}) = {1\over3}$ and $P(\text{tails}) = {2\over3}$. Suppose that the coin will be repeatedly flipped until at least two heads and at least two tails are obtained. Letting $X$ be the number of flips required, give the value of $P(X= 6)$

Attempted Solution:

In order to get the second head on the $6^\text{th}$ toss, you need to get $4$ tails in the first $5$ tosses, and heads on the $6^\text{th}$ toss. Similarly, in order to get the second tail on the $6^\text{th}$ toss, you need to get $4$ heads in the first $5$ tosses, and tails on the $6^\text{th}$ toss. So would the answer just be

$${5\choose4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 + {5\choose4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^2 = 0.1372$$

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  • $\begingroup$ Hm, I'm not sure where I might have gone wrong. $\endgroup$ – Remy Oct 8 '17 at 22:21
  • $\begingroup$ I initially misread part of this. As it is it looks ok. $\endgroup$ – Michael Hardy Oct 8 '17 at 22:26
  • $\begingroup$ Ok great, thanks $\endgroup$ – Remy Oct 8 '17 at 22:27

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