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Let a fair die be rolled 2 times. The 2 rolls are independent. Let X and Y be the outcomes of the first and second rolls, respectively.

a. What is the probability distribution of X+Y? What is each unique possible value of X+Y and each possibility’s corresponding probability.

b. What is the probability that X+Y is greater or equal to 10?

Answer: I previously posted this without making an attempt. Here is my attempt to this:

a) 2 happens 1 time, 3 - 2 times, 4 - 3 times, 5 - 4 times, 6 - 5 times, 7 - 6 times, 8 - 5 times, 9 - 4 times, 10 - 3 times, 11 - 2 times, 12 - 1 time.

so the probability distribution is: 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36 and 1/36, respectively.

b) since 10 happens 3 times, 11 - 2 times, and 12 - 1 time, answer is 6/36.

Anna
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  • The numbers, reasoning, are correct. I would suggest saying that for example there are $3$ ordered pairs $(x,y)$ such that $x+y=4$. Each has probability $\frac{1}{36}$, so $\Pr(X+Y=4)=\frac{3}{36}$. – André Nicolas Jun 15 '14 at 20:21

2 Answers2

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Yes you got it right, congratulations. I really don't know what else to add...

JoeyBF
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All that I can suggest is to express your numbers as a piecewise function:

$$\newcommand{\P}{\operatorname{\mathcal{P}}}\P(X+Y = z) = \begin{cases}(z-1)/36 & \text{if: }2\leq z\leq 7, z\in \mathbb{Z} \\ (13-z)/36 & \text{if: } 7\lt z\leq 12, z\in \mathbb{Z} \\0 & \text{elsewhere} \end{cases}$$

Graham Kemp
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