The conditional probability of an event A, assuming that an event B has occured is defined by
P(A|B)=\frac{P(A\cap B)}{P(B)}
n(S)=6\cdot 6=36 . the set of all equally likely outcomes
A - “The first die shows a 2, a 3 or a 5”
B - “The second die shows an add number”
n(A)=18
n(B)=18
A\cap B =\{(2,1), (2,3),(2,5), (3,1, (3,3), (3,5), (5,1),(5,3),(5,5)\}
n(A\cap B)=9
P(B)=\frac{n(A)}{n(S)}=\frac{18}{36}=\frac{1}{2}
P(A\cap B)=\frac{9}{36}=\frac{1}{4}
P(A|B)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\cdot {\frac{2}{1}}=\frac{1}{2}
Two dice rolls - table
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)