Two biased dice are thrown together. For the first die $P(6)=\frac{1}{2},$ other scores are equally likely. While for the second die, $P(1)=\frac{2}{5}$ and other scores are equally likely then the mean for the probability distribution of " the number of ones seen", will be

$\frac{1}{2}$

The correct answer is Option (4) → $\frac{1}{2}$

$E(X)=E(I_1)+E(I_2)$, where $I_1$ and $I_2$ are variables that equal one if the third and second die show a 1, and 0 otherwise.

First Die

$P(6)=\frac{1}{2}$, Sum of probability of remaining 5 faces = $1-\frac{1}{2}=\frac{1}{2}$

Probability for each of faces other than 6 = $\frac{\frac{1}{2}}{5}=\frac{1}{10}$

Second Die

$P(1)=\frac{2}{5}$, Probability of remaining = $1-\frac{2}{5}=\frac{3}{5}$

Probability of each of faces other than 1 = $\frac{\frac{3}{5}}{5}=\frac{3}{25}$

$E(I_1)$ = P(first die shows 1) = $\frac{1}{10}$

$E(I_2)$ = P(second die shows 1) = $\frac{2}{5}$

$E(X)==E(I_1)+E(I_2)=\frac{1}{10}+\frac{2}{5}=\frac{5}{10}=\frac{1}{2}$