Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.

Explanation

Out of the numbers 1 to 9, there are 4 even numbers and 5 odd numbers.
Let A = event of choosing two odd numbers out of 1 to 9, and
B = event of choosing two numbers whose sum is even
`therefore` n(A) = numbers of ways of choosing 2 odd numbers out of 5 = `""^3C_2`
n(B) = numbers of ways of choosing 2 numbers out of 1 to 9 whose sum is even
`=""^4C_2+""^5C_2`
then, `n(AcapB)` = number of ways of choosing 2 odd numbers whose sum is even = `""^9C_2`
`therefore " " P(B)=(n(B))/(n(S))=(""^4C_2+""^5C_2)/(""^9C_2)`
`=((4xx3)/(2xx1)+(5xx4)/(2xx1))/((9xx8)/(2xx1))=32/72=4/9`
`P(AcapB)=(n(AcapB))/(n(S))=(""^5C_2)/(""^9C_2)=(5xx4)/(9xx8)=5/18`
`therefore` The required probability = P(A|B)
`=(P(AcapB))/(P(B))=(5/18)/(4/9)=5/8`