# Probability of a biased coin obtaining its second heads or seconds tails on the $6^\text{th}$ toss

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

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

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