A coin is biased so that the head is three times as likely to occur as tail. If the coin is tossed twice. The mean of the number of tails is :

$\frac{1}{2}$

Let probability of tail = $p$, probability of head = $3p$.

Since $p+3p=1 \;\;\Rightarrow\;\; 4p=1 \;\;\Rightarrow\;\; p=\frac{1}{4},\;\; P(\text{head})=\frac{3}{4}$

Define random variable $X=$ number of tails in two tosses.

Then $X \sim \text{Binomial}(n=2, p=\frac{1}{4})$

Mean of binomial distribution: $E(X)=np=\;2\times \frac{1}{4}=\frac{1}{2}$

Answer: Mean number of tails $=\;\frac{1}{2}$