When a die is rolled, the sample space is given by
S={1,2,3,4,56}.
Let A = event of getting a prime number, and
B = event of getting and odd number.
Then, A={2,3,5},B={1,3,5} and A∩B={3,5}.
∴P(A)=
n(A)
n(S)
=
3
6
=
1
2
,P(B)=
n(B)
n(S)
=
3
6
=
1
2
and P(A∩B)=
n(A∩B)
n(S)
=
2
6
=
1
3
.
Suppose B has already occurred and then A occurs. So, we have to find P(A/B).
Now, P(A/B)=
P(A∩B)
P(B)
=
(1/3)
(1/2)
=(
1
3
×
2
1
)=
2
3
P(A//B)=(P(A nn B))/(P(B))=((1//3))/((1//2))=(1/3xx2/1)=2/3
Hence, the required probability is 2/3.
