$A$ coin is biased so that the head is $3$ times as likely to occur as tail. If the coin is tossed twice,find the probability distribution of number of tails.

(D) Let the probability of getting a tail in the biased coin be $x$.
$\therefore P(T) = x$
$\Rightarrow P(H) = 3x$
For a biased coin,$P(T) + P(H) = 1$
$\Rightarrow x + 3x = 1$
$\Rightarrow 4x = 1$
$\Rightarrow x = \frac{1}{4}$
$\therefore P(T) = \frac{1}{4}$ and $P(H) = \frac{3}{4}$
When the coin is tossed twice,the sample space is $\{HH, TT, HT, TH\}$.
Let $X$ be the random variable representing the number of tails.
$\therefore P(X=0) = P(\text{no tail}) = P(H) \times P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$
$P(X=1) = P(\text{one tail}) = P(HT) + P(TH) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$
$P(X=2) = P(\text{two tails}) = P(TT) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
Therefore,the required probability distribution is as follows:

$X$	$0$	$1$	$2$
$P(X)$	$\frac{9}{16}$	$\frac{3}{8}$	$\frac{1}{16}$