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13 Jan 2026, 11:19
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John is applying to be the leader of a study group. To qualify, he can either take a short test in which he must solve both of the 2 questions, or he can take a longer test in which he must solve at least 3 of the 4 questions. John’s probability of solving any one question correctly is \(p\), where \(0 < p < 1\). Would John have a better chance of qualifying if he chose the longer test?
(1) The probability that John qualifies if he chooses the short test is less than \(\frac{1}{4}\).
(2) The probability that John does not qualify if he chooses the short test is greater than \(\frac{8}{9}\).
(1) The probability that John qualifies if he chooses the short test is less than \(\frac{1}{4}\).
(2) The probability that John does not qualify if he chooses the short test is greater than \(\frac{8}{9}\).
13 Jan 2026, 11:19
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Official Solution:
John is applying to be the leader of a study group. To qualify, he can either take a short test in which he must solve both of the 2 questions, or he can take a longer test in which he must solve at least 3 of the 4 questions. John’s probability of solving any one question correctly is \(p\), where \(0 < p < 1\). Would John have a better chance of qualifying if he chose the longer test?
Short test qualifying probability is \(p^2\).
Longer test qualifying probability is P(exactly 3 out of 4 correct) + P(all 4 correct) \(= \frac{4!}{3!} * p^3(1 - p) + p^4 = 4p^3 - 3p^4\).
The longer test is better when \(4p^3 - 3p^4 > p^2\):
\(4p - 3p^2 > 1\)
\(3p^2 - 4p + 1 < 0\)
\((3p - 1)(p - 1) < 0\)
\(\frac{1}{3} < p < 1\)
Therefore, the answer to the question is YES when p is greater than \(\frac{1}{3}\), and no when \(p\) is less than or equal to \(\frac{1}{3}\).
(1) The probability that John qualifies if he chooses the short test is less than \(\frac{1}{4}\).
This means \(p^2 < \frac{1}{4}\), so \(p < \frac{1}{2}\). \(p\) could still be below \(\frac{1}{3}\) or above \(\frac{1}{3}\). Not sufficient.
(2) The probability that John does not qualify if he chooses the short test is greater than \(\frac{8}{9}\).
Not qualifying on the short test is \(1 - p^2\). So, we get \(1 - p^2 > \frac{8}{9}\), which gives \(p^2 < \frac{1}{9}\), thus \(p < \frac{1}{3}\). If \(p < \frac{1}{3}\), the longer test is not better. Sufficient.
Answer: B
John is applying to be the leader of a study group. To qualify, he can either take a short test in which he must solve both of the 2 questions, or he can take a longer test in which he must solve at least 3 of the 4 questions. John’s probability of solving any one question correctly is \(p\), where \(0 < p < 1\). Would John have a better chance of qualifying if he chose the longer test?
Short test qualifying probability is \(p^2\).
Longer test qualifying probability is P(exactly 3 out of 4 correct) + P(all 4 correct) \(= \frac{4!}{3!} * p^3(1 - p) + p^4 = 4p^3 - 3p^4\).
The longer test is better when \(4p^3 - 3p^4 > p^2\):
\(4p - 3p^2 > 1\)
\(3p^2 - 4p + 1 < 0\)
\((3p - 1)(p - 1) < 0\)
\(\frac{1}{3} < p < 1\)
Therefore, the answer to the question is YES when p is greater than \(\frac{1}{3}\), and no when \(p\) is less than or equal to \(\frac{1}{3}\).
(1) The probability that John qualifies if he chooses the short test is less than \(\frac{1}{4}\).
This means \(p^2 < \frac{1}{4}\), so \(p < \frac{1}{2}\). \(p\) could still be below \(\frac{1}{3}\) or above \(\frac{1}{3}\). Not sufficient.
(2) The probability that John does not qualify if he chooses the short test is greater than \(\frac{8}{9}\).
Not qualifying on the short test is \(1 - p^2\). So, we get \(1 - p^2 > \frac{8}{9}\), which gives \(p^2 < \frac{1}{9}\), thus \(p < \frac{1}{3}\). If \(p < \frac{1}{3}\), the longer test is not better. Sufficient.
Answer: B
16 Jan 2026, 10:19
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Shouldn't that be p<1/3 or p<1? That is what we do in quadratic right?
Bunuel
