Coin $1$ is a fair coin and coin $2$ is an unfair coin such that the probability of getting heads is $0.6.$ One of the coins is chosen at random and flipped repeatedly until the first head is obtained. Suppose that the first head is observed in the fifth flip. What is the probability that coin $2$ was taken?
Suppose $A$ is the event that indicates coin $1$ was taken and $B$ is the event that indicates coin $2$ was taken. Let $X$ be the random variable that indicates number of flips required to get the first head. Since one of the coins was taken at random so $P(A) = P(B) = \frac 1 2.$ Also by that given condition $P(X=5 \mid B) = (0.4)^4 (0.6).$ So $P(X=5 \mid B^c) = P(X=5 \mid A) = (0.5)^5.$ We need to find out $P(B \mid X=5).$ Now by Bayes' theorem \begin{align*} P(B \mid X=5) & = \frac {P(X=5 \mid B) P(B)} {P(X=5 \mid B) P(B) + P(X=5 \mid B^c) P(B^c)} \\ & = \frac {(0.4)^4 \times (0.6) \times (0.5)} {(0.4)^4 \times (0.6) \times (0.5) + (0.5)^5 \times (0.5)} \\ & = \frac {(0.4)^4 \times (0.6)} {(0.4)^4 \times (0.6) + (0.5)^5} \\ & \simeq 0.33. \end{align*}
Am I correct? Please check my solution.
Thanks in advance.
Yes, your approach and your result are correct.