Out of the numbers from 1 to 9, there are 5 odd numbers and 4 even numbers.
Let A = event of choosing two odd numbers, and B = event of choosing two numbers whose sum is even.
then, n(A)= number of ways of choosing 2 odd numbers out of 5
=.5C2.
n(B)= number of ways of choosing 2 numbers whose sum is even
=(.4C2+.5C2) [2 out of 4 even and 2 out of 5 odd].
n(A∩B)= number of ways of choosing 2 odd numbers out of 5
=.5C2.
Suupose B has already occurred and then A occurs.
and P(A∩B)=
n(A∩B)
n(S)
=
2
36
=
1
18
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)
=
(1/18)
(5/36)
=(
1
18
×
36
5
)=
2
5
.
Hence, the required probability is
2
5
.
