A six faced die is biased such that 3 × P (a prime number) = 6 × P...

Let $\dfrac{P\left(\text{a prime number}\right)}{2} = \dfrac{P\left(\text{a composite number}\right)}{1} = \dfrac{P\left(1\right)}{3} = k$

So, $P$(a prime number)$= 2k$,

$P$ (a composite number)$= k$,

and $P\left(1\right) = 3k$

i.e. $6k + 2k + 3k = 1$

$⇒ k = \dfrac{1}{11}$

$P ( \text{success} ) = P\left(1\text{or}4\right) = 3k + k = \dfrac{4}{11}$

Number of trials, $n = 2$

$∴$ required mean$= np = 2 \times \dfrac{4}{11} = \dfrac{8}{11}$