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15 Mar 2026, 13:43
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (01:50) correct
77%
(01:54)
wrong
based on 53
sessions
History
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A manager assigns five distinct tasks at random to three interns. If each intern is assigned at least one task, what is the probability that one intern is assigned three tasks?
A. 1/5
B. 1/4
C. 1/3
D. 2/5
E. 1/2
A. 1/5
B. 1/4
C. 1/3
D. 2/5
E. 1/2
16 Mar 2026, 01:49
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Since each intern gets at least one task, the only ways to split 5 tasks among 3 interns are: 3-1-1 or 2-2-1. We need to count the arrangements for each split.
Counting 3-1-1 (one intern gets 3 tasks):
- Choose which intern gets 3 tasks: 3 ways
- Choose which 3 of the 5 tasks they get: C(5,3) = 10 ways
- The remaining 2 tasks go one each to the other 2 interns: 2! = 2 ways
- Total = 3 × 10 × 2 = 60
Counting 2-2-1 (one intern gets 1 task, two get 2 each):
- Choose which intern gets just 1 task: 3 ways
- Choose which 1 of the 5 tasks they get: 5 ways
- Split the remaining 4 tasks into 2 groups of 2 for the other 2 interns: C(4,2) = 6 ways (note that no need to divide by 2 since the interns are distinct people)
- Total = 3 × 5 × 6 = 90
Total valid arrangements = 60 + 90 = 150
Probability that one intern gets 3 tasks = 60 / 150 = 2/5
Answer: D
Key Principle: When distributing distinct items to distinct people with constraints, list all valid partitions first, then count each partition carefully by choosing WHO gets the special role, WHICH items they get, and HOW the rest are distributed.
Counting 3-1-1 (one intern gets 3 tasks):
- Choose which intern gets 3 tasks: 3 ways
- Choose which 3 of the 5 tasks they get: C(5,3) = 10 ways
- The remaining 2 tasks go one each to the other 2 interns: 2! = 2 ways
- Total = 3 × 10 × 2 = 60
Counting 2-2-1 (one intern gets 1 task, two get 2 each):
- Choose which intern gets just 1 task: 3 ways
- Choose which 1 of the 5 tasks they get: 5 ways
- Split the remaining 4 tasks into 2 groups of 2 for the other 2 interns: C(4,2) = 6 ways (note that no need to divide by 2 since the interns are distinct people)
- Total = 3 × 5 × 6 = 90
Total valid arrangements = 60 + 90 = 150
Probability that one intern gets 3 tasks = 60 / 150 = 2/5
Answer: D
Key Principle: When distributing distinct items to distinct people with constraints, list all valid partitions first, then count each partition carefully by choosing WHO gets the special role, WHICH items they get, and HOW the rest are distributed.
17 Mar 2026, 02:27
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What a wonderful and detailed explanation! Thanks for taking the time to write this
