You have a fair coin and a biased coin where there is both head. You pick a coin randomly with no knowledge whether it is biased or unbiased. You toss three times . First two tosses yield head. What is the probability of getting head in third toss.

I tried to solve it and my answer is coming 9/10 . Can anyone please verify ?

  • 1
    $\begingroup$ I come up with the same answer. I suggest that you post some of your work / rationale in future posts. $\endgroup$ – Doug M Jan 26 '18 at 19:07

Let $H_i$ be the event 'The $i$-th toss comes up heads' and similarly for $T_i$ with tails. Let $F$ be the event that 'The chosen coin is fair' and $B$ be the event that 'The chosen coin is biased'. Then

\begin{align}\mathbb P(H_3|H_1\cap H_2)=\frac{\mathbb P(H_3\cap H_1\cap H_2)}{\mathbb P(H_1\cap H_2)}\end{align}

Now, we have

\begin{align} \mathbb P(H_3\cap H_1\cap H_2) &=\mathbb P(H_3\cap H_1\cap H_2|F)\cdot \mathbb P(F)+\mathbb P(H_3\cap H_1\cap H_2|B)\cdot \mathbb P(B)\\ &=\frac1{2^3}\cdot\frac12+1\cdot\frac12=\frac1{16}+\frac12=\frac9{16} \end{align}


\begin{align} \mathbb P(H_1\cap H_2) &=\mathbb P(H_1\cap H_2|F)\cdot \mathbb P(F)+\mathbb P( H_1\cap H_2|B)\cdot \mathbb P(B)\\ &=\frac1{2^2}\cdot\frac12+1\cdot\frac12=\frac1{8}+\frac12=\frac5{8}=\frac{10}{16} \end{align}

Hence we have that

\begin{align}\mathbb P(H_3|H_1\cap H_2) &=\frac{\frac9{16}}{\frac{10}{16}}=\frac9{10},\end{align}

as you claimed.


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