A die is biased such that 3P(prime)=6P(composite)=2P(1).

Question

A die is biased such that $3P(	ext{prime})=6P(	ext{composite})=2P(1)$. $X$ counts perfect squares in two throws. Find the mean of $X$.

Accepted Answer

Let $3\alpha=6\beta=2\gamma=k$. With $3\alpha+2\beta+\gamma=1$ (weighted by counts: 3 primes, 2 composites, 1 unity face), $k+\dfrac{k}{3}+\dfrac{k}{2}=1\Rightarrow k=\dfrac{6}{11}$.

$P(\text{perfect square})=P(1)+P(4)=\dfrac{k}{2}+\dfrac{k}{6}=\dfrac{4}{11}$.

$$\text{Mean} = 2\times\frac{4}{11}=\frac{8}{11}$$

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