A six faced die is biased such that 3 × P (a prime number) = 6 × P (a composite number) = 2 ×…

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

So, $P$(a prime number)$=2k$,

$P$ (a composite number)$=k$,

and $P\left(1\right)=3k$

i.e. $6k+2k+3k=1$

$\Rightarrow k=\frac{1}{11}$

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

Number of trials, $n=2$

$\therefore$ required mean$=np=2\times \frac{4}{11}=\frac{8}{11}$