Two biased dice are thrown together. For the first die $P(6) = \frac{1}{2}$,

the other scores being equally likely while for the second die $P(1) = \frac{2}{5}$ and the other scores are equally likely. Find the probability distribution of 'the number of one's seen'

For first die, $P(6) = \frac{1}{2}$

and $P\left( {{6^\prime }} \right) = \frac{1}{2}$
$\Rightarrow$ $P(1) + P(2) + P(3) + P(4) + P(5) = \frac{1}{2}$

$\Rightarrow$ $P(1) = \frac{1}{{10}}$

and $P\left( {{1^\prime }} \right) = \frac{9}{{10}}$

For second die, $P(1) = \frac{2}{5}$

and $P\left( {{1^\prime }} \right) = 1 - \frac{2}{5} = \frac{3}{5}$

Let $X =$ Number of one's seen

For $X = 0,$ $P(X = 0) = P\left( {{1^\prime }} \right) \cdot P\left( {{1^\prime }} \right) = \frac{9}{{10}} \cdot \frac{3}{5} = \frac{{27}}{{50}} = 0.54$

$P(X = 1) = P\left( {{1^\prime }} \right) \cdot P\left( {{1^\prime }} \right) + P\left( {{1^\prime }} \right) \cdot P\left( {{1^\prime }} \right) = \frac{9}{{10}} \cdot \frac{2}{5} + \frac{1}{{10}} \cdot \frac{3}{5}$

$= \frac{{18}}{{50}} + \frac{3}{{50}} = \frac{{21}}{{50}} = 0.42$

$P(X = 2) = P(1) \cdot P(1) = \frac{1}{{10}} \cdot \frac{2}{5} = \frac{2}{{50}} = 0.04$

Hence, the required probability distribution is as below