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22 Jan 2025, 01:30
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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:
35%
(medium)
Question Stats:
75% (01:47) correct
25%
(02:17)
wrong
based on 326
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A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?
A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28
A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28
22 Jan 2025, 02:50
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Total number of ways to select a committee of 3 from 8 employees:
This is a combination problem, so the total number of ways is:
C(8, 3) = 8! / (3! * (8 - 3)!) = 56.
Now, we need to select a committee that includes Alice but not Bob.
1. Alice is already selected, so we need to choose 2 more members from the remaining 6 employees (excluding Alice and Bob).
The number of ways to do this is:
C(6, 2) = 6! / (2! * (6 - 2)!) = 15.
Thus, the probability that the committee includes Alice but not Bob is:
15 / 56.
Answer: D. 15/56
This is a combination problem, so the total number of ways is:
C(8, 3) = 8! / (3! * (8 - 3)!) = 56.
Now, we need to select a committee that includes Alice but not Bob.
1. Alice is already selected, so we need to choose 2 more members from the remaining 6 employees (excluding Alice and Bob).
The number of ways to do this is:
C(6, 2) = 6! / (2! * (6 - 2)!) = 15.
Thus, the probability that the committee includes Alice but not Bob is:
15 / 56.
Answer: D. 15/56
22 Jan 2025, 04:24
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A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?
Total number of committees that can be formed = 8C3
Now the Favorable cases are: Alice should be in the committee and Bob should not be in the committee.
Suppose Alice has been selected to the committee, the remaining two positions can be selected from remaining 7-1 members (As Bob can not be selected).
Hence the number of favorable cases = 6C2
Hence the probability required= 6C2/8C3= 15/56
IMO Ans D
Total number of committees that can be formed = 8C3
Now the Favorable cases are: Alice should be in the committee and Bob should not be in the committee.
Suppose Alice has been selected to the committee, the remaining two positions can be selected from remaining 7-1 members (As Bob can not be selected).
Hence the number of favorable cases = 6C2
Hence the probability required= 6C2/8C3= 15/56
IMO Ans D
