#### 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*}
Before the experiment, we would have said there was only a \(\frac{1}{6}\) chance that the sum of the rolls is at least 10. However, with the additional knowledge of seeing the first roll of 6, we find the probability of a sum of at least 10 to be \(\frac{1}{2}\text{,}\) substantially more likely than before.
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