An unfair 3-sided die is rolled twice. The probability of rolling a 3 is $0.5$, the probability of rolling a 1 is $0.25$, and the probability of rolling a 2 is $0.25$. Let $X$ be the outcome of the first roll and $Y$ the outcome of the second.

  • Find the Joint Distribution of $X$ and $Y$ in a Table.

    The outcome of $X = \{1,2,3\}$.

    The outcome of $Y = \{1,2,3\}$.

    Would I just make a table of all the roll possibilities?

  • Find the Probability $\mathrm{P}(X+Y \geq 5)$.

    The only roll that will make this is a 3 or a 2. Should I just take the same of every possible roll to find this probability?

  • $\begingroup$ What does a three-sided die look like? $\endgroup$ – amWhy Feb 16 '15 at 17:38
  • $\begingroup$ Just imagine a dice with 3 sides only :D Or a 6-sided dice with 3 sides' probability being 0 $\endgroup$ – Gareth Ma Feb 26 '18 at 23:34
  • $\begingroup$ Latest reply after 2 years $\endgroup$ – Gareth Ma Feb 26 '18 at 23:34

Sounds as though you are very much on the right track with this computation. Yes, make a table of all roll possibilities, and in each entry of the table (e.g. $X = i, Y = j$) find the probability of that outcome (since you are rolling two separate times, you can treat $X$ and $Y$ as independent random variables). Once you have your table, it will be easy to total up the probabilities for the outcomes which meet the condition $X+Y \ge 5$ (you can use the fact that the different outcomes which satisfy this condition are mutually exclusive).

  • $\begingroup$ Yes I got that far, however, I am confused with this part. When X = 1 and Y = 1, would the probability be 0.25 or the product of the two probabilities? $\endgroup$ – Reed Rogaski Feb 16 '15 at 18:00
  • $\begingroup$ @ReedRogaski The probability would be $0,25(0,25)$ since the two rolls are independent of each other. $\endgroup$ – grayQuant Feb 16 '15 at 18:07

Cool. I answered like this

$\begin{array}{c|c:c:c|c}X\backslash Y& 1&2&3\\\hline 1&1/16& 1/16& 1/8&1/4\\\hdashline 2 & 1/16& 1/16& 1/8&1/4\\\hdashline 3 & 1/8& 1/8& 1/4&1/2\\\hline&1/4&1/4&1/2&1\end{array}$

Marginals both sum to 1.

Probability is $\mathsf P((X,Y)\in\{(3,2), (2,3), (3,3)\})$ $= 1/8 + 1/8 + 1/4 \\= 1/2$

Does that look right?


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