1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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