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 $X$.


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Graham Kemp, José Carlos Santos, hardmath, JonMark Perry, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

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


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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.