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

Question

A six faced die is biased such that

$$3 \times \mathrm{P}($$a prime number$$)\,=6 \times \mathrm{P}($$a composite number$$)\,=2 \times \mathrm{P}(1)$$.

Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :

Accepted Answer

Let P(a prime number) = $$\alpha$$

P(a composite number) = $$\beta$$

and P(1) = $$\gamma$$

$$\because$$ $$3\alpha  = 6\beta  = 2\gamma  = k$$ (say)

and $$3\alpha  + 2\beta  + \gamma  = 1$$

$$ \Rightarrow k + {k \over 3} + {k \over 2} = 1 \Rightarrow k = {6 \over {11}}$$

Mean = np where n = 2

and p = probability of getting perfect square

$$ = P(1) + P(4) = {k \over 2} + {k \over 6} = {4 \over {11}}$$

So, mean $$ = 2\,.\,\left( {{4 \over {11}}} \right) = {8 \over {11}}$$

Solutions · 1