Question

A box contains 10 coins, 4 of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1 / 2$, while it is $2 / 3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. If the probability that the coin drawn is fair can be expressed as rational $\frac{a}{b}$, where a and b are relatively prime, then find the value of $(a+b)$.

Answer 1

$A =$ event what we have $=$ coins drawn tossed twice, found to fall as HT
$B _1=$ It is a fair coins,
$P \left( B _1\right)=\frac{4}{10}=\frac{2}{5}$
$B _2= It$ is a biased coins, $\quad P \left( B _2\right)=\left(1-\frac{2}{5}\right)=\frac{3}{5}$
$P \left( A / B _1\right)=\frac{1}{4}, \quad P \left( A / B _2\right)=\frac{2}{9}$
$P \left( B _1 / A \right)=\frac{ P \left( B _1\right) \cdot P \left( A / B _1\right)}{ P \left( B _1\right) \cdot P \left( A / B _1\right)+ P \left( B _2\right) \cdot P \left( A / B _2\right)}=\frac{\frac{2}{5} \cdot \frac{1}{4}}{\frac{2}{5} \cdot \frac{1}{4}+\frac{3}{5} \cdot \frac{2}{9}}$
$=\frac{\frac{1}{10}}{\frac{1}{10}+\frac{2}{15}}=\frac{3}{3+4}=\frac{3}{7}=\frac{ a }{ b }$
Hence $a+b=10$