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16 Aug 2024, 01:30
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C
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Difficulty:
65%
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Question Stats:
60% (02:30) correct
41%
(02:53)
wrong
based on 200
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A 4-person committee is to be formed from a group of 6 Democrats and 5 Republicans. Which of the following is the closest approximation to the probability that the committee will consist of at least 3 Democrats?
A. 0.11
B. 0.22
C. 0.35
D. 0.42
E. 0.75
A. 0.11
B. 0.22
C. 0.35
D. 0.42
E. 0.75
16 Aug 2024, 02:54
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IMO C
There are two favorable outcomes, either (a) 3 democrats and 1 republican are selected or (b) 4 democrats and 0 republicans are selected.
The probability for (a) is given by (6C3 * 5C1) / 11C4 i.e., 10/33.
The probability for (b) is given by 6C4 / 11C4 i.e., 1/22.
We sum the two to get 23/66 or ~0.35.
There are two favorable outcomes, either (a) 3 democrats and 1 republican are selected or (b) 4 democrats and 0 republicans are selected.
The probability for (a) is given by (6C3 * 5C1) / 11C4 i.e., 10/33.
The probability for (b) is given by 6C4 / 11C4 i.e., 1/22.
We sum the two to get 23/66 or ~0.35.
16 Aug 2024, 12:33
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Total ways to select a 4 person committee = 11C4 = 330
Atleast 3 Dem => Exactly 3 Dem and 1 Rep (6C3 * 5C1) OR All 4 Dem (6C4)
Probability of at least 3 Dem = (6C3 * 5C1 + 6C4) / 330 => 23/66 i.e. pretty close to 1/3 (0.33) therefore answer ~ C.) 0.35
Atleast 3 Dem => Exactly 3 Dem and 1 Rep (6C3 * 5C1) OR All 4 Dem (6C4)
Probability of at least 3 Dem = (6C3 * 5C1 + 6C4) / 330 => 23/66 i.e. pretty close to 1/3 (0.33) therefore answer ~ C.) 0.35
