Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Let X is the random variable score obtained when a die is thrown twice.

$\therefore$ $X = 1,2,3,4,5,6$

Here, $S = \{ (1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3), \ldots ,(6,6)\}$

$\therefore$ $P(X = 1) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{{36}}$

$P(X = 2) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{3}{{36}}$

$P(X = 3) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{5}{{36}}$

Similarly, $P(X = 4) = \frac{7}{{36}}$

$P(X = 5) = \frac{9}{{36}}$

$P(X = 6) = \frac{{11}}{{36}}$

So, the required distribution is,

Also, we know that,

Mean $\{ E(X)\} = \Sigma XP(X)$
$= \frac{1}{{36}} + \frac{6}{{36}} + \frac{{15}}{{36}} + \frac{{28}}{{36}} + \frac{{45}}{{36}} + \frac{{66}}{{36}} = \frac{{161}}{{36}}$