Question

A bag contains $10$ coins of which $5$ are normal coins, $2$ of them doubly-headed coins and $3$ of them are weighted coins with heads occuring $3$ times as probable as tails. A coin is randomly taken out from the bag and is tossed for three times, the coin shows heads on all the three occasions. If the probability that the coin was a normal coin is $\frac{m}{n},$ where $m$ and $n$ are co-prime, then $n-m$ is

Answer 1

$A:$ Coin tossed $3$ times, shows heads on all $3$ times.
$B_1:$ It is a normal coin; $P\left(B_1\right)=\frac{5}{10}$
and $P\left(\text{heads}\right)=\frac{1}{2}$
$B_2:$ It is a doubly-headed coin; $P\left(B_2\right)=\frac{2}{10}$
and $P\left(\text{heads}\right)=1$
$B_3:$ It is a weighted coin; $P\left(B_3\right)=\frac{3}{10}$
and $P\left(\text{heads}\right)=\frac{3}{4}$

$P\left(A|B_1\right)=\frac{1}{8};P\left(A|B_2\right)=1;P\left(A|B_3\right)=\frac{27}{64}$

Now, $P\left(B_1|A\right)=\frac{P(B_1\cap A)}{P\left(A\right)}$
$=\frac{\frac{5}{10}.\frac{8}{64}}{\frac{5}{10}.\frac{8}{64}+\frac{2}{10}.\frac{64}{64}+\frac{3}{10}.\frac{27}{64}}$
$=\frac{40}{40+128+81}=\frac{40}{249}$