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
- $194/285$
- $1/57$
- $13/19$
- $3/4$
The total number of ways to choose 3 integers from 20 is C(20, 3) = 1140. The product is NOT a multiple of 3 if all three chosen numbers are not multiples of 3. There are 6 multiples of 3 (3, 6, 9, 12, 15, 18) and 14 non-multiples. The number of ways to pick 3 from the 14 non-multiples is C(14, 3) = 364. The probability of the product NOT being a multiple of 3 is 364/1140 = 91/285. The probability of it being a multiple is 1 - 91/285 = 194/285.
To find the probability that the product is a multiple of 3, it is easier to use the complement by finding the probability that none of the 3 chosen numbers is a multiple of 3. There are 6 multiples of 3 in the first 20 natural numbers, leaving 14 numbers that are not multiples of 3. The number of ways to choose 3 numbers from these 14 is (14 choose 3), which equals 364, and the total ways to choose any 3 numbers from 20 is (20 choose 3), which equals 1140. The probability of not getting a multiple of 3 is 364/1140, so subtracting this from 1 gives 776/1140, which simplifies to 194/285.