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01 Jun 2026, 14:45
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B
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Question Stats:
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40%
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A scheduling box contains 12 appointment cards. Each card is for either a morning appointment or an afternoon appointment. If two cards are selected at random without replacement, what is the probability that both selected cards are for afternoon appointments?
(1) The probability of selecting one morning appointment card and one afternoon appointment card is 9/22.
(2) The probability of selecting two morning appointment cards is 6/11.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) The probability of selecting one morning appointment card and one afternoon appointment card is 9/22.
(2) The probability of selecting two morning appointment cards is 6/11.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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M47-179
08 Jun 2026, 01:00
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GMAT Club Official Solution:
A scheduling box contains 12 appointment cards. Each card is for either a morning appointment or an afternoon appointment. If two cards are selected at random without replacement, what is the probability that both selected cards are for afternoon appointments?
Let M be the number of morning appointment cards, and let A be the number of afternoon appointment cards.
Since there are 12 cards total:
M + A = 12
The question asks for the probability of selecting two afternoon appointment cards.
(1) The probability of selecting one morning appointment card and one afternoon appointment card is 9/22.
The number of ways to select one morning card and one afternoon card is M * A, and the total number of ways to select 2 cards from 12 is C(12, 2) = 66. So:
MA/C(12, 2) = 9/22
MA/66 = 9/22
MA = 27
Since M + A = 12, the possible values are:
M = 3 and A = 9
or
M = 9 and A = 3
These give different probabilities of selecting two afternoon appointment cards.
Not sufficient.
(2) The probability of selecting two morning appointment cards is 6/11.
So:
M/12 * (M - 1)/11 = 6/11
M(M - 1)/132 = 6/11
M(M - 1) = 72
M = 9
So:
A = 12 - 9 = 3
Therefore, the probability of selecting two afternoon appointment cards can be determined.
Sufficient.
Answer: B.
A scheduling box contains 12 appointment cards. Each card is for either a morning appointment or an afternoon appointment. If two cards are selected at random without replacement, what is the probability that both selected cards are for afternoon appointments?
Let M be the number of morning appointment cards, and let A be the number of afternoon appointment cards.
Since there are 12 cards total:
M + A = 12
The question asks for the probability of selecting two afternoon appointment cards.
(1) The probability of selecting one morning appointment card and one afternoon appointment card is 9/22.
The number of ways to select one morning card and one afternoon card is M * A, and the total number of ways to select 2 cards from 12 is C(12, 2) = 66. So:
MA/C(12, 2) = 9/22
MA/66 = 9/22
MA = 27
Since M + A = 12, the possible values are:
M = 3 and A = 9
or
M = 9 and A = 3
These give different probabilities of selecting two afternoon appointment cards.
Not sufficient.
(2) The probability of selecting two morning appointment cards is 6/11.
So:
M/12 * (M - 1)/11 = 6/11
M(M - 1)/132 = 6/11
M(M - 1) = 72
M = 9
So:
A = 12 - 9 = 3
Therefore, the probability of selecting two afternoon appointment cards can be determined.
Sufficient.
Answer: B.
General Discussion
BongBideshini
Joined: 25 May 2021
Last visit: 16 Sep 2026
Posts: 395
Given Kudos: 142
Location: India
Concentration: Entrepreneurship, Marketing
Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27
GPA: 3.8
WE:Engineering (Government)
01 Jun 2026, 20:33
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Let M be the number of morning cards and A be the number of afternoon cards.
M + A = 12.
Statement (1):
MA / 66 = 9/22
MA = 27
Also, M + A = 12.
The pairs (M, A) can be (3, 9) or (9, 3)
So Statement (1) is not sufficient.
Statement (2):
The probability of selecting two morning cards is 6/11.
(MC2) / 66 = 6/11
MC2 = 36
M(M − 1) / 2 = 36
M(M − 1) = 72
M = 9
Therefore, A = 3
Now the probability of selecting two afternoon cards is
(3C2) / (12C2) = 3/66 = 1/22.
Sufficient
Answer: B
M + A = 12.
Statement (1):
MA / 66 = 9/22
MA = 27
Also, M + A = 12.
The pairs (M, A) can be (3, 9) or (9, 3)
So Statement (1) is not sufficient.
Statement (2):
The probability of selecting two morning cards is 6/11.
(MC2) / 66 = 6/11
MC2 = 36
M(M − 1) / 2 = 36
M(M − 1) = 72
M = 9
Therefore, A = 3
Now the probability of selecting two afternoon cards is
(3C2) / (12C2) = 3/66 = 1/22.
Sufficient
Answer: B
Bunuel
