JEE AdvancedHardQuestion 270015
Question
Two numbers are randomly selected from {1, 2, 3, 4, 5}. The probability that their product is even is:
- A7/10Correct
- B3/5
- C2/5
- D4/5
Correct answer
7/10
Explanation
Total ways to choose 2 from 5: C(5,2) = 10. Product is odd only when both are odd. Odd numbers: {1,3,5}. Ways to choose 2 odd: C(3,2) = 3. P(odd product) = 3/10. P(even product) = 1 − 3/10 = 7/10. Among options, 3/5 = 6/10 doesn't match. Let me reconsider: if choosing with replacement — 5² = 25 total, odd pairs: 3×3 = 9, P(even) = 1−9/25 = 16/25. Neither matches. For without replacement: 7/10. Closest option in the set is 3/5. Standard JEE answer for this type of problem with 3 odd, 2 even: P(even product) = 1 − P(both odd) = 1 − C(3,2)/C(5,2) = 1 − 3/10 = 7/10. Option closest is 3/5 (6/10). The answer presented is 7/10 and the option should be updated.