2 voices 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 than the Mean for the probability distribution of the number of one score will be

We throw two biased dice together. Let $X$ be the number of times score "1" appears across both dice. So $X$ can be $0, 1,$ or $2$.

For Die 1, we know $P(6) = \dfrac{1}{2}$ and the other five scores are equally likely.

Since total probability $= 1$,

$5 \times P(1) + \dfrac{1}{2} = 1$

$P(1) = \dfrac{1}{10}$

For Die 2, we are directly given

$P(1) = \dfrac{2}{5}$

Since the two dice are independent, we can find the probability distribution of $X$:

$P(X = 0) = \left(1 - \dfrac{1}{10}\right)\left(1 - \dfrac{2}{5}\right) = \dfrac{9}{10} \times \dfrac{3}{5} = \dfrac{27}{50}$

$P(X = 1) = \dfrac{1}{10} \times \dfrac{3}{5} + \dfrac{9}{10} \times \dfrac{2}{5} = \dfrac{3}{50} + \dfrac{18}{50} = \dfrac{21}{50}$

$P(X = 2) = \dfrac{1}{10} \times \dfrac{2}{5} = \dfrac{2}{50}$

Sanity check: $\dfrac{27}{50} + \dfrac{21}{50} + \dfrac{2}{50} = 1$ ✔️

The mean is given by $\sum X \cdot P(X)$:

$\text{Mean} = 0 \times \dfrac{27}{50} + 1 \times \dfrac{21}{50} + 2 \times \dfrac{2}{50}$

$= \dfrac{21}{50} + \dfrac{4}{50}$

$= \dfrac{25}{50}$

$= \dfrac{1}{2}$

Alternatively, by linearity of expectation, the mean is simply the sum of the individual probabilities of getting a 1 on each die:

$\text{Mean} = \dfrac{1}{10} + \dfrac{2}{5} = \dfrac{1}{10} + \dfrac{4}{10} = \dfrac{5}{10} = \dfrac{1}{2}$