Chapter 2 Probability - 2.4 Conditional Probability - Questions - Page 38: 13

The probability that the number on the first die was at least as large as 4 given that the sum of the two dice was 8 is 3/5.

We express two die being tossed as: $S=\left\{ \begin{align} & \left( 1,\text{ }1 \right)\text{ }\left( 1,\text{ }2 \right)\text{ }\left( 1,\text{ }3 \right)\text{ }\left( 1,\text{ }4 \right)\text{ }\left( 1,\text{ }5 \right)\text{ }\left( 1,\text{ }6 \right)\text{ } \\ & \left( 2,\text{ }1 \right)\text{ }\left( 2,\text{ }2 \right)\text{ }\left( 2,\text{ }3 \right)\text{ }\left( 2,\text{ }4 \right)\text{ }\left( 2,\text{ }5 \right)\text{ }\left( 2,\text{ }6 \right)\text{ } \\ & \left( 3,\text{ }1 \right)\text{ }\left( 3,\text{ }2 \right)\text{ }\left( 3,\text{ }3 \right)\text{ }\left( 3,\text{ }4 \right)\text{ }\left( 3,\text{ }5 \right)\text{ }\left( 3,\text{ }6 \right)\text{ } \\ & \left( 4,\text{ }1 \right)\text{ }\left( 4,\text{ }2 \right)\text{ }\left( 4,\text{ }3 \right)\text{ }\left( 4,\text{ }4 \right)\text{ }\left( 4,\text{ }5 \right)\text{ }\left( 4,\text{ }6 \right)\text{ } \\ & \left( 5,\text{ }1 \right)\text{ }\left( 5,\text{ }2 \right)\text{ }\left( 5,\text{ }3 \right)\text{ }\left( 5,\text{ }4 \right)\text{ }\left( 5,\text{ }5 \right)\text{ }\left( 5,\text{ }6 \right)\text{ } \\ & \left( 6,\text{ }1 \right)\text{ }\left( 6,\text{ }2 \right)\text{ }\left( 6,\text{ }3 \right)\text{ }\left( 6,\text{ }4 \right)\text{ }\left( 6,\text{ }5 \right)\text{ }\left( 6,\text{ }6 \right) \\ \end{align} \right\}$ Let A be the event that the number on the first die was at least as large as 4 and B be the even that the sum of the two dice was 8. We see that 15 of the sample outcomes in S constitute the event A: $A=\left\{ \begin{align} & \left( 4,\text{ }4 \right)\text{ }\left( 4,\text{ }5 \right)\text{ }\left( 4,\text{ }6 \right)\text{ }\left( 5,\text{ }1 \right)\text{ }\left( 5,\text{ }2 \right)\text{ } \\ & \left( 5,\text{ }3 \right)\text{ }\left( 5,\text{ }4 \right)\text{ }\left( 5,\text{ }5 \right)\text{ }\left( 5,\text{ }6 \right)\text{ }\left( 6,\text{ }1 \right)\text{ } \\ & \left( 6,\text{ }2 \right)\text{ }\left( 6,\text{ }3 \right)\text{ }\left( 6,\text{ }4 \right)\text{ }\left( 6,\text{ }5 \right)\text{ }\left( 6,\text{ }6 \right) \\ \end{align} \right\}$ \[P(A)=\frac{15}{36}\] We see that 5 of the sample outcomes in S constitute the event B: $B=\{\left( 2,\text{ }6 \right)\text{ }\left( 3,\text{ }5 \right)\text{ }\left( 4,\text{ }4 \right)\text{ }\left( 5,\text{ }3 \right)\text{ }\left( 6,\text{ }2 \right)\}$ \[P(B)=\frac{5}{36}\] $A\cap B=\{\left( 4,\text{ }4 \right)\text{ }\left( 5,\text{ }3 \right)\text{ }\left( 6,\text{ }2 \right)\}$ \[P(A\cap B)=\frac{3}{36}\] We have to find the probability of that the number on the first die was at least as large as 4 given that the sum of the two dice was 8 as: \[\begin{align} & P(A|B)=\frac{P(A\cap B)}{P(B)} \\ & P(A|B)=\frac{{}^{3}/{}_{36}}{{}^{5}/{}_{36}} \\ & P(A|B)=\frac{3}{5} \\ \end{align}\] Therefore, the probability of that the number on the first die was at least as large as 4 given that the sum of the two dice was 8 is 3/5.