Question

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up head $75\%$ of the time and third is an unbiased coin. One of the three coins is choosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer 1

Let
$A$: be an event that two headed coin is choosen
$B$: be an event that biased coin is choosen
$C$: be an event that unbiased coin is choosen
$H$: be an event that head appears on the coin

CASE-1

Probability that two headed coin is choosen
$P\left(A\right)=\frac{1}{3}$ ...$\left(1\right)$

Probability that head appear on the two headed coin.
$P\left(H|A\right)=1$ ...$\left(2\right)$

CASE-2

Probability that biased coin is choosen
$P\left(B\right)=\frac{1}{3}$ ...$\left(3\right)$

Probability that head appear on the biased coin.
$P\left(H|B\right)=75\%=\frac{75}{100}=\frac{3}{4}$ ....$\left(4\right)$

CASE-3

Probability that unbiased coin is choosen
$P\left(C\right)=\frac{1}{3}$ ...$\left(5\right)$

Probability that head appear on the unbiased coin.
$P\left(H|C\right)=\frac{1}{2}$ ...$\left(6\right)$

By Bayes' theorem,

$P\left(A|H\right)=\left(\frac{P\left(A\right)\times P\left(H|A\right)}{P\left(A\right)\times P\left(H|A\right)+P\left(B\right)\times P\left(H|B\right)+P\left(C\right)\times P\left(H|C\right)}\right)$ ...$\left(7\right)$

Substituting the value of$\left(1\right),\left(2\right),\left(3\right),\left(4\right),\left(5\right)\&\left(6\right)\text{in}\left(7\right)$,

$P\left(A|H\right)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{3}\times \frac{1}{2}}$

$P\left(A|H\right)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}|1+\frac{3}{4}+\frac{1}{2}|}$

$P\left(A|H\right)=\frac{1}{\frac{9}{4}}=\frac{4}{9}$

Therefore, the probability that coin is two headed if it shows head =$\frac{4}{9}$.