Question

There are three coins.One is a two headed coin,another is a biased coin that comes up heads $75\%$ of the and third is an unbiased coin.One of the three coin is chosen at random and tossed,it shows head,what is the probability that it was the two headedcoin?

Answer 1

The correct answer is: $\frac{4}{9}$
Let $E_1=$a two-headed coin, $E_2=$a biased coin and $A=$a head is shown
Now $P(E_1)=\frac{1}{3}, P(E_2)=\frac{1}{3}, P(E_3)=\frac{1}{3}$
$P(A|E_1)=1, P(A|E_2)=\frac{75}{100}=\frac{3}{4}$ and $P(A|E_3)=\frac{1}{2}$
Therefore, by Bayes' theorem,
$P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}$
$=\frac{\frac{1}{3}×1}{\frac{1}{3}×1+\frac{1}{3}×\frac{3}{4}+\frac{1}{3}×\frac{1}{2}}$
$=\frac{\frac{1}{3}}{\frac{1}{3}(1+\frac{3}{4}+\frac{1}{2})}$
$=\frac{4}{9}$