A biased coin in which the head is 3 times as likely to occur as tail, is tossed twice. If the probability distribution of the number of tails is given by:

The value of $\frac{m}{n}$ is:

6

Let x be the random variable as ‘Number of tails’.

∴ P.D table

∴ x = 0, 1, 2

x = 0 means no tail

x = 1 means 1 tail

x = 2 means both tail

$P(H)=\frac{3}{4}$, $P(T)=\frac{1}{4}$

S = {HH, HT, TH, TT}

P(x = 0) = P(HH) = $\frac{3}{4}.\frac{3}{4}=\frac{9}{16}$

P(x = 1) = P(HT or TH) = P(HT) + P(TH) = $\frac{3}{4}.\frac{1}{4}+\frac{1}{4}.\frac{3}{4}=\frac{6}{16}=\frac{3}{8}$

P(x = 2) = P(TT) = $\frac{1}{4}.\frac{1}{4}=\frac{1}{16}$

∵ P(x = 1) = m = $\frac{3}{8}$

& P(x = 2) = n = $\frac{1}{16}$

$\frac{m}{n}=\frac{\frac{3}{8}}{\frac{1}{16}}$

$=\frac{3}{8}×\frac{16}{1}=6$