We have 25 numbers in a row, one after another, without skipping any numbers. The smallest number in this set is an odd number.
This means there are 13 odd numbers and 12 even numbers among them. (Odd and even numbers always alternate. If we start with an odd number, we will always have one more odd number than even numbers.)
We need to find the chance (probability) that if we pick 3 of these numbers, their total (sum) is an even number.
For the sum of 3 numbers to be even, two situations can happen:
1. We choose three even numbers. (Even + Even + Even
= Even)
2. We choose two odd numbers and one even number. (Odd + Odd + Even
= Even)
Counting the Ways
Number of ways to pick 3 even numbers:
There are 12 even numbers. Number of ways
=‌12C3.
Number of ways to pick 2 odd numbers and 1 even number:
There are 13 odd numbers and 12 even numbers.
Ways to pick 2 odd numbers
=‌13C2.
Ways to pick 1 even number
=‌12C1.
Total ways
=‌13C2×‌12C1.
Total Possible Ways
The total number of ways to select any 3 numbers from 25 numbers is
‌25C3.
Probability Formula
The probability that the sum is even is:
P(A)=‌| ‌13C2⋅‌12C1+‌12C3 |
| ‌25C3 |
=‌
