A coin is baised 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.

.: Let $X$ denote the random variable which

represents the number of tails on the baised coin tossed twice.

$\therefore$ $X$ can assume values 0, 1, 2

let $p$ be the probability of success $= \cfrac{1}{4}$

$q$ be the probability of failure $= \cfrac{3}{4}$

$\therefore$ $P(X = 0) =$ Probability that both heads appear

$= q \times q = \cfrac{3}{4} \times \cfrac{3}{4} = \cfrac{9}{{16}}$

$P(X = 1) =$ Probability that one head and one tail appears

$= 2(p \times q) = 2\left( {\cfrac{1}{4} \times \cfrac{3}{4}} \right) = \cfrac{3}{8}$

and $P(X = 2) =$ Probability that both tails appear

$= p \times p = \cfrac{1}{4} \times \cfrac{1}{4} = \cfrac{1}{{16}}$