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

Answer: 0013

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
$\therefore P(E)=P\left(E_1\right)\cdot P\left(E/E_1\right)+P\left(E_2\right)\cdot P\left(E/E_2\right)+P\left(E_3\right)\cdot P\left(E/E_3\right)$

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