Question

Three distinct integers are chosen at random from the first 20 positive integers. Compute the probability that: (a) their sum is even; (b) their product is even.

Short answer

The probability that their sum is even is \(\frac{1}{2}\) and the probability that their product is even is \(\frac{85}{95}\).

Step-by-step solution

Total number of ways to pick 3 numbers out of 20

First, let's calculate the total number of ways to pick 3 numbers out of 20. This is done using the combination formula: \(_{20}C_{3} = \frac{20!}{[(20-3)!*3!]} = 1140\) ways.

Find number of ways for three integers to have even sum

The sum of three integers is even if all of them are even or two of them are odd (because odd + odd = even and then summing this with another even number gives an overall even sum). Let's calculate each case separately: there are \(_{10}C_{3} = 120\) ways to choose all even (from the 10 even numbers out of 20), and \(_{10}C_{2} = 45\) ways to choose 2 odd numbers (from the 10 odd numbers), then multiplied by 10 ways to choose 1 even number, which gives 450 ways. So, in total, there are \(120 + 450 = 570\) ways.

Probability that the sum of these integers is even

Probability is calculated as the ratio of the favorable outcomes over the total number of outcomes. Here, the number of favorable outcomes is 570 (from step 2), and total outcomes is 1140 (from step 1). Hence, the required probability = \(\frac{570}{1140} = \frac{1}{2}\).

Number of ways for the product of three integers to be even

The product of any set of numbers is even if at least one of the numbers is even. So, it's easier to find the ways where all numbers are odd, and subtract that from the total. There are \(_{10}C_{3} = 120\) ways to choose all odd numbers (from the 10 odd numbers).

Probability that the product of these integers is even

The cases where their product is even will be obtained by subtracting the all-odd case from the total. That is, \(1140 - 120 = 1020\) ways. Hence, the required probability = \(\frac{1020}{1140} = \frac{85}{95}\), after simplifying.