There are nine pong balls numbered 1-9 in a bag If three balls are randomly selected without replacement what is the probability expressed as a common fraction that the sum of the number be odd?

Answer 1

10/21

First, observe that for the sum of the three balls to be odd, either one is odd and two are even, or all three are odd. Since the sum of two odd numbers is even, the sum of two even number is even, and the sum of an even number and an odd number is odd.

P(exactly one is odd)=(5/9)*(4/8)*(3/7)*3=180/504=5/14

P(all three are odd)=(5/9)*(4/8)*(3/7)=60/504=5/42

P(sum is odd)=5/14+5/42=15/42+5/42=20/42=10/21, which is approximately .48