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28 Sep 2021, 09:16
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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 \(\frac{2}{31}\) ?
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than \(\frac{5}{6}\).
(2) If the die is rolled twice, the probability that the numbers obtained are equal is greater than \(\frac{1}{8}\).
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than \(\frac{5}{6}\).
(2) If the die is rolled twice, the probability that the numbers obtained are equal is greater than \(\frac{1}{8}\).
28 Sep 2021, 09:16
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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
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
24 May 2023, 05:07
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Can somebody explain this?
P(two different numbers)=1−P(two same numbers)=1−1∗1n. How is probability of 2 same nos. is 1/n. Shouldnt it be 1/n *1/n ?
P(two different numbers)=1−P(two same numbers)=1−1∗1n. How is probability of 2 same nos. is 1/n. Shouldnt it be 1/n *1/n ?
