venksune wrote:

1. Total no. of ways of choosing 2 numbers out of the 10 numbers = 10C2 = 45.

2. The choice of the two numbers shouldbe such that one of them have to be even and the other number can be any of the remaining numbers -so that the prod of tjhe numbers will be even.

We have 2, 4, 6, 8 as the even numbers. Selecting one among them is 4C1 ways = 4 ways.

Having selected one even number we can choose the other number from any of the remaining 9 numbers in 9C1 or 9 ways. Remember 0 is a even number.

The probability is thus (4*9)/45 = 4/5

couple of things venksune

first off nowhere is it mentioned that numbers cant be repeated

so I think the total number od cases is 10* 10 = 100

another thing is you considered the case where the first number of the product is even , but what about the condition where the 2nd number is even

so I think the toatl number of favourable cases are 4*9 + 9*4 = 72

so the final ans is 72/100

Your thoughts??