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A fair six-sided die carries $1$ on one face, $2$ on two of its faces, and
$3$ on the remaining three faces.

Suppose the die is rolled twice, and let $X$ be the random variable ’total score'. Find the probability distribution of $X$.

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closed as off-topic by Graham Kemp, José Carlos Santos, hardmath, JonMark Perry, Trevor Gunn Oct 18 '17 at 15:59

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  • $\begingroup$ You are expected to show some of your own work. What have you tried so far, and what is causing you trouble? $\endgroup$ – Graham Kemp Oct 17 '17 at 16:14
  • $\begingroup$ Getting the probability distribution is the major problem here.But finding the variance and mean is not a problem $\endgroup$ – joshua mwakio Oct 17 '17 at 16:17
  • $\begingroup$ Why is it a problem? You have a support of only five values whose probability mass is readily apparent. $\endgroup$ – Graham Kemp Oct 17 '17 at 16:25
  • $\begingroup$ I dont know how to distribute its probability mass $\endgroup$ – joshua mwakio Oct 17 '17 at 16:27
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    $\begingroup$ Well, what is $\mathsf P(T=t)$ for each $t\in\{2,3,4,5,6\}$? That is it. $\endgroup$ – Graham Kemp Oct 17 '17 at 16:29
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The six-sided die has 1 on one face, 2 on two faces, and 3 on three faces.   That gives the support and priobabilities for an individual roll.

The die is rolled twice, and the result of each roll added to give $T$.   Thus let $T=T_1+T_2$, with $T_1,T_2$ being the individual die rolls, which are independent and identically distributed as above.

$$\therefore\qquad \mathsf P(T{=}t) ~=~\sum_{s=\max\{1,\,t-3\}}^{\min\{3,\,t-1\}} \mathsf P(T_1{=}s)\,\mathsf P(T_2{=}t{-}s) ~\mathbf 1_{t\in\{2,3,4,5,6\}}$$

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For example, $P(T=2)$ is the probability that we get $1$ at both tries. Since the trials are independent, this means:

$P(T=2) = P(1~on~first~try)\times P(1~on~second~try)=1/6\times 1/6=1/36$. You can work through all other situations as it is suggested.

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