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23 Aug 2021, 08:00
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B
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An n-sided fair die has sides labeled with the numbers 1 through n, inclusive, where n > 3. If the die is rolled once, is the probability of rolling a "8" greater than 2/31 ?
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than 5/6.
(1) If the die is rolled twice, the probability that the numbers obtained are equal is greater than 1/8.
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than 5/6.
(1) If the die is rolled twice, the probability that the numbers obtained are equal is greater than 1/8.
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28 Sep 2021, 09:17
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Bunuel
Official Solution:
An \(n\)-sided fair die has sides labeled with the numbers 1 through \(n\), inclusive, where \(n > 3\). If the die is rolled once, is the probability of rolling a "8" greater than \(\frac{2}{31}\) ?
The probability of rolling a "8" when rolling an \(n\)-sided die is \(\frac{1}{n}\), provided \(n\) is greater than or equal to 8 (if \(n < 8\), then the probability of rolling a "8" will be 0). So, the question asks whether \(\frac{1}{n} > \frac{2}{31}\), (which translates into \(n < 15.5\)) AND \(n > 7\). Thus the question became: is \(7 < n < 15.5\)
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than \(\frac{5}{6}\).
\(P(two \ different \ numbers) = 1 - P(two \ same \ numbers)=1-1*\frac{1}{n}\). So, we are told that \(1-\frac{1}{n} > \frac{5}{6}\):
\(1-\frac{1}{n} > \frac{5}{6}\);
\(\frac{1}{6} > \frac{1}{n} \);
\(n>6\) (since \(n\) is positive).
If \(n=7\), then the probability of rolling a "8" is 0, so less than \(\frac{2}{31}\).
If \(8 \leq n \leq 15\), then the probability of rolling a "8" will be more than \(\frac{2}{31}\).
And if \(n > 15\), then the probability of rolling a "8" will be less than \(\frac{2}{31}\).
Not sufficient.
(2) If the die is rolled twice, the probability that the numbers obtained are equal is greater than \(\frac{1}{8}\).
\(P(two \ equal \ numbers)=n*\frac{1}{n}*\frac{1}{n}=\frac{1}{n} > \frac{1}{8}\). This leads to \(n < 8\). Therefore, the probability of rolling a "8" is 0. Sufficient.
Answer: B
General Discussion
23 Aug 2021, 08:28
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My answer is B).
We do not know the value of n (other than n>3), so depending on the value of n, there could be 2 cases:
1) if 3<n<8, then Pr(8) = 0
2) if n>=8, then Pr(8) = 1/n
We want to know if Pr(8)>2/31, which is equivalent to 1/n>2/31 -> n<31/2=15.5. Since n is an integer, 8<=n<=15 is the only range of value of n when this would be possible.
Statement 1: the probability of obtaining the same #s after 2 roles is = n*1/n*1/n = 1/n. So 1-1/n>5/6 -> 1/n<1/6 -> n>6. Insufficient because n could be 9 which falls into [8,15], or n could be 30.
Statement 2: from the above, 1/n>1/8 -> n<8. We know definitively that n would be outside of the range of [8,15], so the answer is no. Sufficient.
B)
We do not know the value of n (other than n>3), so depending on the value of n, there could be 2 cases:
1) if 3<n<8, then Pr(8) = 0
2) if n>=8, then Pr(8) = 1/n
We want to know if Pr(8)>2/31, which is equivalent to 1/n>2/31 -> n<31/2=15.5. Since n is an integer, 8<=n<=15 is the only range of value of n when this would be possible.
Statement 1: the probability of obtaining the same #s after 2 roles is = n*1/n*1/n = 1/n. So 1-1/n>5/6 -> 1/n<1/6 -> n>6. Insufficient because n could be 9 which falls into [8,15], or n could be 30.
Statement 2: from the above, 1/n>1/8 -> n<8. We know definitively that n would be outside of the range of [8,15], so the answer is no. Sufficient.
B)
