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03 Aug 2026, 10:33
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At an animal shelter, 12 animals were available for adoption. Some were dogs, and the rest were cats. If two animals are selected at random, without replacement, is the probability that both selected animals are dogs greater than \(\frac{1}{3}\)?
(1) Fewer than half of the animals were cats.
(2) The probability that one dog and one cat are selected is \(\frac{16}{33}\).
(1) Fewer than half of the animals were cats.
(2) The probability that one dog and one cat are selected is \(\frac{16}{33}\).
03 Aug 2026, 10:33
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Official Solution:
At an animal shelter, 12 animals were available for adoption. Some were dogs, and the rest were cats. If two animals are selected at random, without replacement, is the probability that both selected animals are dogs greater than \(\frac{1}{3}\)?
Let \(d\) be the number of dogs. Then the question asks:
Is \(\frac{d}{12} * \frac{d - 1}{11} > \frac{1}{3}\)?
Is \(d(d - 1) > 44\)?
By trial and error we can get that this is equivalent to asking whether \(d > 7\).
(1) Fewer than half of the animals were cats.
This means the number of cats was less than 6. Thus, \(d > 6\).
Not sufficient.
(2) The probability that one dog and one cat are selected is \(\frac{16}{33}\).
The probability of selecting one dog and one cat is
\(\frac{d}{12} * \frac{12 - d}{11} * 2! = \frac{16}{33}\)
\(d(12 - d) = 32\)
This gives \(d = 4\) or \(d = 8\). One gives a NO answer, the other gives a YES answer.
Not sufficient.
(1) + (2) From (1), \(d > 6\), so from (2), \(d\) can only be 8. Therefore, the answer is yes. Sufficient.
Answer: C
At an animal shelter, 12 animals were available for adoption. Some were dogs, and the rest were cats. If two animals are selected at random, without replacement, is the probability that both selected animals are dogs greater than \(\frac{1}{3}\)?
Let \(d\) be the number of dogs. Then the question asks:
Is \(\frac{d}{12} * \frac{d - 1}{11} > \frac{1}{3}\)?
Is \(d(d - 1) > 44\)?
By trial and error we can get that this is equivalent to asking whether \(d > 7\).
(1) Fewer than half of the animals were cats.
This means the number of cats was less than 6. Thus, \(d > 6\).
Not sufficient.
(2) The probability that one dog and one cat are selected is \(\frac{16}{33}\).
The probability of selecting one dog and one cat is
\(\frac{d}{12} * \frac{12 - d}{11} * 2! = \frac{16}{33}\)
\(d(12 - d) = 32\)
This gives \(d = 4\) or \(d = 8\). One gives a NO answer, the other gives a YES answer.
Not sufficient.
(1) + (2) From (1), \(d > 6\), so from (2), \(d\) can only be 8. Therefore, the answer is yes. Sufficient.
Answer: C
