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Let X be the sum of the two values and let Y be the product of the two values.

What is the value of P(Y = i) for i = 1,2,3...36.

I am having trouble approaching this problem. We are learning about random variables and bernoullis, geometric etc...

Any help would be appreciated

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What are the "values"? – Sanath K. Devalapurkar Mar 11 '14 at 0:01
The values are the sum of the two dice (1,1) (1,2) ...(6,6) – user134547 Mar 11 '14 at 0:05
You mean $a+a$ or the probability of getting $(a,a)$? – Sanath K. Devalapurkar Mar 11 '14 at 0:06
What is the probability of values of the dice adding up to values between 1 and 12. Example, (1,6) (2,5) (3,4) (6,1) (5,2) (4,3) all add up to 7, so would the probability for 7 just be 1/36? – user134547 Mar 11 '14 at 0:11
Oh. Ok - I understand now. – Sanath K. Devalapurkar Mar 11 '14 at 0:12

Hint: When you throw two six-sided dice, you can have $6\times 6=36$ possible results that are equally likely, that is each has a probability of $1/36$. The results can be represented as pairs of the form $(x,y)$, for example $(1,1), (1,2), (2,1)$ etc. Then the probability of each given event is calculated as a fraction where the numerator is the number of favourable events and the denominator the number of possible events, which is always $36$ (in this exercise). For example the probability that the sum of the dice is equal to $4$, i.e. the probability $P(X=4)$ is calculated as follows $$P(X=4)=\frac{\text{number of results with sum 4}}{\text{number of possible results}}=\frac{\left|(1,3),(2,2),(3,1)\right|}{36}=\frac{3}{36}$$ So list the 36 results and then calculate the favourable number of events for each event that you are asked. A practical way to list all 36 results is the following $$Ω=\pmatrix{(1,1), (1,2), \ldots, (1,6)\\(2,1), (2,2), \ldots, (2,6)\\ \ldots \\(6,1), (6,2), \ldots, (6,6)\\ }$$ that is in a matrix with six rows (first dice) and six columns (second dice).

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+1, fantastic explanation. – recursive recursion Mar 11 '14 at 0:27
@recursiverecursion Thank you for your comment! – Jimmy R. Mar 11 '14 at 0:32

Hint: In general, since each outcome (possible dice roll, like $(3,4)$ or $(5,5)$) is equally likely, we have $$P(X = i) = \frac{\text{number of outcomes where the sum of the dice is $i$}}{\text{total number of possible outcomes}}.$$

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