Chapter 39 - Wave Functions and Uncertainty - Exercises and Problems - Page 1174: 5

a) $\dfrac{1}{6}$ b) $\dfrac{1}{6}$ c) $\dfrac{5}{18}$

Here is a table of all possible outcomes of rolling two dice. \begin{array}{|c|c|} \hline \text{A} & \text{B} \\ \hline 1 & 1 \\ \hline 1 & 2 \\ \hline 1 & 3 \\ \hline 1 & 4 \\ \hline 1 & 5 \\ \hline 1 & 6 \\ \hline 2 & 1 \\ \hline 2 & 2 \\ \hline 2 & 3 \\ \hline 2 & 4 \\ \hline 2 & 5 \\ \hline 2 & 6 \\ \hline 3 & 1 \\ \hline 3 & 2 \\ \hline 3 & 3 \\ \hline 3 & 4 \\ \hline 3 & 5 \\ \hline 3 & 6 \\ \hline 4 & 1 \\ \hline 4 & 2 \\ \hline 4 & 3 \\ \hline 4 & 4 \\ \hline 4 & 5 \\ \hline 4 & 6 \\ \hline 5 & 1 \\ \hline 5 & 2 \\ \hline 5 & 3 \\ \hline 5 & 4 \\ \hline 5 & 5 \\ \hline 5 & 6 \\ \hline 6 & 1 \\ \hline 6 & 2 \\ \hline 6 & 3 \\ \hline 6 & 4 \\ \hline 6 & 5 \\ \hline 6 & 6 \\ \hline \end{array} From the table, we can see that there are 36 possible outcomes. $$\color{blue}{\bf [a]}$$ From the table above, the ways of rolling a 7 are $(1,6),(2,5), (3,4), (4,2), (5,2),(6,1)$ so the probability of rolling a 7 is $$P_{\rm 7}=\dfrac{6}{36} =\color{red}{\bf \dfrac{1}{6}}$$ $$\color{blue}{\bf [b]}$$ A double occurs when both dice show the same number. From the table, the doubles are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ so the probability of rolling any double is $$P_{\rm double}=\dfrac{6}{36} =\color{red}{\bf \dfrac{1}{6}}$$ $$\color{blue}{\bf [c]}$$ From the table above, the ways of rolling a 6 or an 8 are as follows For 6: $(1, 5), (2, 4),(3, 3),(4, 2), (5, 1)$ For 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ so the probability of rolling a 6 or an 8 is $$P_{\rm double}=\dfrac{10}{36} =\color{red}{\bf \dfrac{5}{18}}$$