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06 Aug 2026, 11:38
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A quality-control bin contains 30 sensor chips, some of which are faulty. If two chips are selected at random without replacement, is the probability that both selected chips are faulty less than \(\frac{1}{4}\)?
(1) More than 40% of the chips in the bin are faulty.
(2) More than 50% of the chips in the bin are not faulty.
(1) More than 40% of the chips in the bin are faulty.
(2) More than 50% of the chips in the bin are not faulty.
06 Aug 2026, 11:38
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Official Solution:
A quality-control bin contains 30 sensor chips, some of which are faulty. If two chips are selected at random without replacement, is the probability that both selected chips are faulty less than \(\frac{1}{4}\)?
Let \(F\) be the number of faulty chips. The probability that both selected chips are faulty is:
\(\frac{F}{30} * \frac{F - 1}{29} = \frac{F(F - 1)}{30 * 29}\)
The question asks:
Is \(\frac{F(F - 1)}{30 * 29} < \frac{1}{4}\)?
Is \(F(F - 1) < 217.5\)?
By trial and error near the cutoff, if \(F = 15\), then:
\(F(F - 1) = 15 * 14 = 210\)
which is less than 217.5.
If \(F = 16\), then:
\(F(F - 1) = 16 * 15 = 240\)
which is greater than 217.5.
So the question is equivalent to asking whether \(F \leq 15\).
(1) More than 40% of the chips in the bin are faulty.
\(40\%\) of 30 is 12, so:
\(F > 12\)
If \(F = 13\), then \(F \leq 15\), so the answer is YES.
If \(F = 16\), then \(F > 15\), so the answer is NO.
Not sufficient.
(2) More than 50% of the chips in the bin are not faulty.
More than \(50\%\) of 30 means that at least 16 chips are not faulty. Thus, at most 14 chips are faulty:
\(F \leq 14\)
Since \(F \leq 14\), we know that \(F \leq 15\). Therefore, the probability that both selected chips are faulty is less than \(\frac{1}{4}\).
Sufficient.
Answer: B
A quality-control bin contains 30 sensor chips, some of which are faulty. If two chips are selected at random without replacement, is the probability that both selected chips are faulty less than \(\frac{1}{4}\)?
Let \(F\) be the number of faulty chips. The probability that both selected chips are faulty is:
\(\frac{F}{30} * \frac{F - 1}{29} = \frac{F(F - 1)}{30 * 29}\)
The question asks:
Is \(\frac{F(F - 1)}{30 * 29} < \frac{1}{4}\)?
Is \(F(F - 1) < 217.5\)?
By trial and error near the cutoff, if \(F = 15\), then:
\(F(F - 1) = 15 * 14 = 210\)
which is less than 217.5.
If \(F = 16\), then:
\(F(F - 1) = 16 * 15 = 240\)
which is greater than 217.5.
So the question is equivalent to asking whether \(F \leq 15\).
(1) More than 40% of the chips in the bin are faulty.
\(40\%\) of 30 is 12, so:
\(F > 12\)
If \(F = 13\), then \(F \leq 15\), so the answer is YES.
If \(F = 16\), then \(F > 15\), so the answer is NO.
Not sufficient.
(2) More than 50% of the chips in the bin are not faulty.
More than \(50\%\) of 30 means that at least 16 chips are not faulty. Thus, at most 14 chips are faulty:
\(F \leq 14\)
Since \(F \leq 14\), we know that \(F \leq 15\). Therefore, the probability that both selected chips are faulty is less than \(\frac{1}{4}\).
Sufficient.
Answer: B
