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I roll a fair die twice and obtain two numbers: $X_{1}=$ result of the first roll, and $X_{2}=$ result of the second roll. Write down the sample space $S$, and assuming that all outcomes are equally likely (because the die is fair), find the probability of the event $A$ defined as the event that $X_{1}+X_{2}=8$
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The sample space $S$ can be written as
As we see there are $|S|=36$ elements in $S$. To find probability of $A_{1}$ all we need to do is find $M=|A|$. In particular, $A$ is defined as
\begin{aligned} A &=\left\{\left(X_{1}, X_{2}\right) \mid X_{1}+X_{2}=8, X_{1}, X_{2} \in\{1,2, \cdots, 6\}\right\} \\ &=\{(2,6),(3,5),(4,4),(5,3),(6,2)\} . \end{aligned}
Thus, $|A|=5$, which means that
$$P(A)=\frac{|A|}{|S|}=\frac{5}{36}$$

A very common mistake is not distinguishing between, say $(2,6)$ and $(6,2) .$ It is important to note that these are two different outcomes: $(2,6)$ means that the first roll is a 2 and the second roll is a 6, while $(6,2)$ means that the first roll is a 6 and the second roll is a $2 .$ Note that it is very common to write $P\left(X_{1}+X_{2}=8\right)$ when referring to $P(A)$ as defined above. In fact, $X_{1}$ and $X_{2}$ are examples of random variables.
by Diamond (55,179 points)

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