Question

Out of 20 consecutive positive integers, two are chosen at random. The probability that their sum is odd is

• A. \(\frac{10}{19}\)
• B. \(\frac{1}{20}\)
• C. \(\frac{19}{20}\)
• D. \(\frac{9}{19}\)

Answer 1

The Correct Option is A

You have to choose Odd-Even or Even-Odd.
The probabilities of both events are the same: at the start there’s the same amount of odds and evens.
So calculate for one of them and multiply by 22.

First number being odd: \(\frac{10}{20}=\frac{1}{2}\)

Second number being even: \(\frac{10}{19}\)

Overall \(=\frac{10}{19}\times2\times2=\frac{10}{19}\)...
So the correct option is (A)