Question

A bag contains 3 biased coins $B_1, B_2$ and $B_3$ whose probabilities of falling head wise are $1 / 3$, $2 / 3$ and $3 / 4$ respectively. A coin is drawn randomly and tossed, fell head wise. If the probability that the same coin when tossed again will fall head wise is $\frac{m}{n}$, then find the value of $(n-m)$.

Answer 1

A: coin shows up head
$P \left( A / B _1\right)=\frac{1}{3} ; P \left( A / B _2\right)=\frac{2}{3} ; P \left( A / B _3\right)=\frac{3}{4}$
hence $\sum P \left( A / B _{ i }\right)=\frac{7}{4}$
now $ P \left( B _1 / A \right)=\frac{1}{3} \cdot \frac{4}{7}=\frac{4}{21} ; P \left( B _2 / A \right)=\frac{2}{3} \cdot \frac{4}{7}=\frac{8}{21}$ and $P \left( B _3 / A \right)=\frac{3}{4} \cdot \frac{4}{7}=\frac{9}{21}$
Let $B _1 / A = E _1 ; B _2 / A = E _2$ and $B _3 / A = E _3$
$E$ : same coin when tossed will fall heads
$\therefore P(E)= P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right) $

$ =\frac{4}{21} \cdot \frac{1}{3}+\frac{8}{21} \cdot \frac{2}{3}+\frac{9}{21} \cdot \frac{3}{4} $
$ =\frac{16+64+81}{21 \cdot 3 \cdot 4}=\frac{161}{252}=\frac{23}{36}=\frac{m}{n} \Rightarrow n-m=13$