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You have three coins, one with two heads, and two fair. Pick one at random, flip it twice, and get two heads. What is the probability that tossing again I get heads?

I see it this way: call $2H$ the event of tossing two heads and $3H$ the event of getting the third one, $F$ and $R$ the events of picking a fair coin or the rigged one, respectively. Determine $\mathbb P(3H|2H)$. Is this what they are asking?

I go from the basics: $$\mathbb P(3H|2H) = \frac{\mathbb P(3H\cap 2H)}{\mathbb P(2H)}=\frac{\mathbb P(3H\cap 2H \cap F)+\mathbb P(3H\cap 2H \cap R)}{\mathbb P(2H|F)\mathbb P(F)+\mathbb P(2H|R)\mathbb P(R)}.$$ Take $\mathbb P(F)=\frac23$, $\mathbb P(R)=\frac13$, $\mathbb P(2H|F)=\frac14$, $\mathbb P(2H|R)=1$, $\mathbb P(3H\cap 2H \cap F)=\mathbb P(3H \cap F) = \mathbb P(3H | F)\mathbb P(F)= \frac18\frac 23$,$\mathbb P(3H\cap 2H \cap R)=\mathbb P(3H | R) P(R)= \frac13$, and I get $$\mathbb P(3H|2H) = \frac{\frac1{12}+\frac13}{\frac16+\frac13}=\frac56.$$ Is this reasonable?

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$\frac{5}{6}$ looks correct. Since $3H \subset 2H$, using

$$\mathbb P(3H\mid 2H)= \dfrac{\mathbb P(3H\mid F)\mathbb P(F)+\mathbb P(3H\mid R)\mathbb P(R)}{\mathbb P(2H\mid F)\mathbb P(F)+\mathbb P(2H\mid R)\mathbb P(R)} =\dfrac{\frac18\times\frac23+1 \times \frac13}{\frac14\times\frac23+1 \times \frac13}=\frac{5}{6}$$

might have been slightly faster.

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  • $\begingroup$ right, this is faster. Anyway, a quick Monte Carlo converges to 5/6. $\endgroup$
    – marco
    Oct 16, 2020 at 15:18

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