# Biased coin two flips

We have two coins fair and biased. For which the probability of having a head is 2/3. We select a coin at random and then flip it twice. What is the probability that in both flips we have an identical result?

My attempt:

$$\frac{1}{2}\times\frac{1}{2}^4+\frac{1}{2}\times\frac{1}{3}^2\times\frac{2}{3}^2$$

Is my solution correct?

You have the right pieces, but you’ve not put them together correctly. Suppose that you pick the biassed coin: the probability of getting two heads is $\left(\frac23\right)^2$, and the probability of getting two tails is $\left(\frac13\right)^2$, but this means that the probability of getting the same result twice is $\left(\frac23\right)^2+\left(\frac13\right)^2$, not $\left(\frac23\right)^2\cdot\left(\frac23\right)^2$. Either of the outcomes two heads and two tails is a success (meaning both the same), so the probability of a success is the sum of the individual probabilities. You would multiply if you needed both of these things to happen simultaneously – for instance, if you were flipping the coin four times and needed the first two flips to be heads and the last two to be tails.
Similarly, the probability of getting two heads with the fair coin is $\left(\frac12\right)^2$, and so is the probability of getting two tails, so the probability of getting the same result twice is $\left(\frac12\right)^2+\left(\frac12\right)^2$, not $\left(\frac12\right)^4$. Since the probability of picking each coin is $\frac12$, the correct final result is
$$\frac12\left(\left(\frac23\right)^2+\left(\frac13\right)^2\right)+\frac12\left(\left(\frac12\right)^2+\left(\frac12\right)^2\right)\;,$$
$$\frac12\cdot\frac59+\frac12\cdot\frac12=\frac5{18}+\frac14=\frac{19}{36}\;.$$
The probability that the fair coin selected and both flips were identical is $\displaystyle \frac{1}{2}\left(\underbrace{\frac{1}{4}}_{\text{two heads}}+\underbrace{\frac{1}{4}}_{\text{two tails}}\right)$. The probability that the biased coin selected and both flips were identical is $\displaystyle \frac{1}{2}\left(\underbrace{\frac{4}{9}}_{\text{two heads}}+\underbrace{\frac{1}{9}}_{\text{two tails}}\right)$. In total we have $$P=\frac{1}{2}\cdot{\frac{1}{2}}+\frac{1}{2}\cdot{\frac{5}{9}}=\frac{19}{36}$$