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06 Aug 2026, 05:18
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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 \(\frac{9}{22}\).
(2) The probability of selecting two morning appointment cards is \(\frac{6}{11}\).
(1) The probability of selecting one morning appointment card and one afternoon appointment card is \(\frac{9}{22}\).
(2) The probability of selecting two morning appointment cards is \(\frac{6}{11}\).
06 Aug 2026, 05:18
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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 \(\frac{9}{22}\).
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:
\(\frac{MA}{C(12, 2)} = \frac{9}{22}\)
\(\frac{MA}{66} = \frac{9}{22}\)
\(MA = 27\)
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 \(\frac{6}{11}\).
So:
\(\frac{M}{12} * \frac{M - 1}{11} = \frac{6}{11}\)
\(\frac{M(M - 1)}{132} = \frac{6}{11}\)
\(M(M - 1) = 72\)
\(M = 9\)
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 \(\frac{9}{22}\).
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:
\(\frac{MA}{C(12, 2)} = \frac{9}{22}\)
\(\frac{MA}{66} = \frac{9}{22}\)
\(MA = 27\)
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 \(\frac{6}{11}\).
So:
\(\frac{M}{12} * \frac{M - 1}{11} = \frac{6}{11}\)
\(\frac{M(M - 1)}{132} = \frac{6}{11}\)
\(M(M - 1) = 72\)
\(M = 9\)
So:
\(A = 12 - 9 = 3\)
Therefore, the probability of selecting two afternoon appointment cards can be determined.
Sufficient.
Answer: B
