#### Example 1.3.2.

An experiment consists of rolling a fair 6-sided die two times. Let \(A\) be the event that the sum of the rolls is at least 10. To find \(\Pr(A)\text{,}\) we note that \(A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}\text{,}\) which has 6 elements. Since \(|\Omega| = 36\text{,}\) we have
\begin{gather*} \Pr(A) = \frac{|A|}{|\Omega|} = \frac{6}{36} = \frac{1}{6}\text{.} \end{gather*}
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Let \(B\) be the event that the first roll is 6. If we pause after the first die roll seeing the value of 6, we might be more inclined to expect a sum of at least 10. The evidence that we’ve already seen changes our understanding of the situation. Since \(|B| = 6\text{,}\) \(\Pr(B) = \frac{6}{36} = \frac{1}{6}\text{.}\) Also, \(A\cap B = \{(6, 4), (6, 5), (6, 6)\}\text{,}\) so \(\Pr(A\cap B) = \frac{3}{36} = \frac{1}{12}\text{.}\) Finally:
\begin{gather*} \Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{1/12}{1/6} = \frac{1}{2} \end{gather*}
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in-context