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Answer
Two dice are thrown $n=36$
$\begin{array}{l}
S=\{(1,1),(1,2),(1,3),(1,4),(1,5) \text {, } \\
(1,6),(2,1),(2,2),(2,3),(2,4) \text {, } \\
(2,5),(2,6),(3,1),(3,2),(3,3) \text {, } \\
(3,4),(3,5),(3,6),(4,1),(4,2) \text {, } \\
(4,3),(4,4),(4,5),(4,6),(5,1) \text {, } \\
(5,2),(5,3),(5,4),(5,5),(5,6) \text {, } \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}
\end{array}$
Event A: Two numbers appearing on throwing two dice are different.
$\begin{array}{l}
\therefore \quad r=30 \\
\therefore \quad P(A)=\frac{30}{36} \\
=\frac{5}{6}
\end{array}$
Event B: The sum of numbers on the dice is 4 .
$\begin{array}{l}
B=\{(1,3),(2,2),(3,1)\} \\
A \cap B=\{(1,3),(3,1)\} \\
\therefore \quad r=2 \\
\therefore \quad P(A \cap B)=\frac{2}{36}=\frac{1}{18} \\
\therefore \quad P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\
=\frac{\frac{1}{18}}{\frac{5}{6}} \\
=\frac{1}{15}
\end{array}$
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