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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 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.
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M47-227
25 Jul 2026, 02:00
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GMAT Club 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 1/4?
Let F be the number of faulty chips. The probability that both selected chips are faulty is:
F/30 * (F - 1)/29 = F(F - 1)/(30 * 29)
The question asks:
Is F(F - 1)/(30 * 29) < 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 this is equivalent to asking whether F <= 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 <= 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 is more than 15, so at least 16 chips are not faulty. Thus, at most 14 chips are faulty, so F <= 14.
Since F <= 14, then the answer to the question whether F <= 15 is YES.
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 1/4?
Let F be the number of faulty chips. The probability that both selected chips are faulty is:
F/30 * (F - 1)/29 = F(F - 1)/(30 * 29)
The question asks:
Is F(F - 1)/(30 * 29) < 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 this is equivalent to asking whether F <= 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 <= 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 is more than 15, so at least 16 chips are not faulty. Thus, at most 14 chips are faulty, so F <= 14.
Since F <= 14, then the answer to the question whether F <= 15 is YES.
Sufficient.
Answer: B.
General Discussion
21 Jul 2026, 23:49
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Okay this one's sneaky because it looks like a straight probability problem but it's really testing whether you can bound an inequality across a whole range of values, classic Data Sufficiency behavior.
1. Let F be the number of faulty chips out of 30. P(both faulty) = F/30 x (F-1)/29. We want to know if this is less than 1/4.
2. Statement 1 says F > 12, so F could be anywhere from 13 to 30. Check the extremes: at F=13, P is about 13/30 x 12/29, roughly 0.179, under 1/4. At F=30 (all faulty), P = 1, way over 1/4. Since the answer flips depending on where F actually lands, statement 1 alone can't tell you anything for sure. Not sufficient.
3. Statement 2 says more than 50% are NOT faulty, so F is at most 14. Check the worst case, the highest F allowed, F=14: P is 14/30 x 13/29, about 0.209, still under 1/4. And P only grows as F grows, so if the max case is safely under 1/4, every smaller F is too. The answer to the question stays a consistent yes across the whole range. Sufficient.
4. Answer is B.
The trap here is assuming that because statement 1 gives you a number (F>12) and statement 2 gives you a number (F is at most 14), they're somehow equally informative. They're not. Statement 1 puts F in a big range where the probability actually crosses the 1/4 threshold partway through, so you genuinely can't answer yes or no. Statement 2 happens to cap F low enough that the probability never gets close to 1/4 anywhere in its range. I got burned by problems like this early on, assuming any bound on a variable is good enough for sufficiency, when what actually matters is whether the answer stays constant across the entire range that bound allows.
Test the extremes of each statement's range before deciding sufficiency, don't just eyeball the direction of the inequality.
1. Let F be the number of faulty chips out of 30. P(both faulty) = F/30 x (F-1)/29. We want to know if this is less than 1/4.
2. Statement 1 says F > 12, so F could be anywhere from 13 to 30. Check the extremes: at F=13, P is about 13/30 x 12/29, roughly 0.179, under 1/4. At F=30 (all faulty), P = 1, way over 1/4. Since the answer flips depending on where F actually lands, statement 1 alone can't tell you anything for sure. Not sufficient.
3. Statement 2 says more than 50% are NOT faulty, so F is at most 14. Check the worst case, the highest F allowed, F=14: P is 14/30 x 13/29, about 0.209, still under 1/4. And P only grows as F grows, so if the max case is safely under 1/4, every smaller F is too. The answer to the question stays a consistent yes across the whole range. Sufficient.
4. Answer is B.
The trap here is assuming that because statement 1 gives you a number (F>12) and statement 2 gives you a number (F is at most 14), they're somehow equally informative. They're not. Statement 1 puts F in a big range where the probability actually crosses the 1/4 threshold partway through, so you genuinely can't answer yes or no. Statement 2 happens to cap F low enough that the probability never gets close to 1/4 anywhere in its range. I got burned by problems like this early on, assuming any bound on a variable is good enough for sufficiency, when what actually matters is whether the answer stays constant across the entire range that bound allows.
Test the extremes of each statement's range before deciding sufficiency, don't just eyeball the direction of the inequality.
