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Chapter 1: Problem 13

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

Expert verified
The probability that their sum is even is approximately 0.2105 and that their product is even is approximately 0.8947.

Step by step solution

01

Determine the total number of ways three integers can be chosen

This can be calculated by using combination formula \(\binom{n}{k} =\frac{n!}{k!(n-k)!}\). Here, \(n\) is the total number of positives integers, which are 20 in this case, and \(k\) is the number of integers to be chosen, which is 3 in this case, resulting in \(\binom{20}{3} = 1140\) total ways.
02

Calculate the probability that the sum is even

For three integers to have an even sum, all three must be even or all three must be odd. The first 20 positive integers consist of 10 even and 10 odd numbers. We can use the combination formula as in Step 1 to find both cases and add them together to find the total number of ways. The number of ways to pick 3 even numbers from 10 is \(\binom{10}{3} = 120\), and similarly the number of ways to pick 3 odd numbers from 10 is also \(\binom{10}{3} = 120\), therefore in total there are \(120+120 = 240\) ways. Hence, the probability is \(\frac{240}{1140} = 0.2105\) when rounded to 4 decimal places.
03

Calculate the probability that the product is even

For three integers to have an even product, at least one of them must be even. We can find the number of ways where all three numbers are odd and subtract it from the total number of ways. The number of ways to pick 3 odd numbers, similar to step 2, is \(\binom{10}{3} = 120\). Therefore, there are \(1140 - 120 = 1020\) ways where at least one number is even. Hence, the probability is \(\frac{1020}{1140} = 0.8947\) when rounded to 4 decimal places.

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