Three integers are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3 is `194//285` b. `1//57` c. `13//19` d. `3//4`

Options

A. `194//285`
B. `1//57`
C. `13//19`
D. `3//4`

Answer

A

Explanation

The total number of ways of selecting 3 integers from 20 natural numbers is `.^(20)C_(3)` = 1140. Their product is a multiple of 3 means at least one number is divisible by 3. The numbers which are divisible by 3 are 3, 6, 9, 12, 15, 18 and the number of ways of selecting at least one of them is `.^(6)C_(1) xx .^(14)C_(2) + .^(6)C_(2) xx .^(14)C_(1) + .^(6)C_(3) = 776`. Hence, the required probability is `776//1140` = `194//285`.