Question:
Out of 20 consecutive positive integers, two are chosen at random. The probability that their sum is odd is
Out of 20 consecutive positive integers, two are chosen at random. The probability that their sum is odd is
Updated On: Jan 30, 2025
- \(\frac{10}{19}\)
- \(\frac{1}{20}\)
- \(\frac{19}{20}\)
- \(\frac{9}{19}\)
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The Correct Option is A
Solution and Explanation
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)
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