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

Probability distribution:

Mean: $\frac{1}{2}$

The correct answer is Option (1) → 

Probability distribution:

Mean: $\frac{1}{2}$

The coin is so biased that the head is 3 times likely to occur as tail. If $p$ is the probability of getting a tail, then probability of getting a head = $3p$,

so $p+3p=1⇒ 4p=1⇒p=\frac{1}{4}$,

so $q=1-\frac{1}{4}=\frac{3}{4}$

The coin is tossed twice, events are independent, therefore, it is a problem of binomial distribution with $p =\frac{1}{4}, q=\frac{3}{4}$ and $n = 2$.

If X denotes the number of tails, then X can take values 0, 1, 2.

$P(0) = {^2C}_0 q^2 = 1. (\frac{3}{4})^2=\frac{9}{16}$

$P(1) = {^2C}_1 pq = 2.\frac{1}{4}.\frac{3}{4}=\frac{6}{16}$ and

$P(2) = {^2C}_2 p^2 = 1. (\frac{1}{4})^2=\frac{1}{16}$

∴ The probability distribution of the number of tails is $\begin{pmatrix}0&1&2\\\frac{9}{16}&\frac{6}{16}&\frac{1}{16}\end{pmatrix}$

Mean = $Σp_ix_i=\frac{9}{16}×0+\frac{6}{16}×1+\frac{1}{16}×2=\frac{8}{16}=\frac{1}{2}$