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Mrs. Claus is selecting three reindeer from the herd for a special sleigh ride. Among the herd are Rudolph, Blitzen, and Comet. What is the probability that she chooses Rudolph, but not Blitzen and not Comet?
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is 1/14.
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is 5/21.
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is 1/14.
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is 5/21.
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25 Apr 2025, 05:53
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Official Solution:
Mrs. Claus is selecting three reindeer from the herd for a special sleigh ride. Among the herd are Rudolph, Blitzen, and Comet. What is the probability that she chooses Rudolph, but not Blitzen and not Comet?
Assuming there are a total of \(n\) reindeer, the probability that Mrs. Claus chooses Rudolph but not Blitzen or Comet, that is, \(P(Rudolph, X, X)\) where \(X\) stands for any reindeer other than Blitzen and Comet, is:
\(3 * \frac{1}{n} * \frac{n-3}{n-1} * \frac{n-4}{n-2}\)
We multiply by 3 because Rudolph, X, X can be chosen in three different ways: {Rudolph, X, X}; {X, Rudolph, X}; and {X, X, Rudolph}.
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is \(\frac{1}{14}\).
\(P(Rudolph, \ Blitzen, \ X) = 3! * \frac{1}{n} * \frac{1}{n-1} * \frac{n-3}{n-2} = \frac{1}{14}\)
No need to solve this equation. Start substituting values of \(n\) greater than 3 and observe that as \(n\) increases, the value of the left-hand side decreases. Therefore, there is only one value of \(n\) greater than or equal to 4 for which the left-hand side is exactly \(\frac{1}{14}\) (this happens when \(n=9\)). Knowing \(n\) allows us to compute the required probability. Sufficient.
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is \(\frac{5}{21}\).
This implies:
\(P(X, X, X) = \frac{n-3}{n} * \frac{n-4}{n-1} * \frac{n-5}{n-2} = \frac{5}{21}\)
Similarly, there is no need to solve this equation. \(n\) must be at least 6, and as \(n\) increases, the left-hand side increases toward 1. Therefore, there is only one value of \(n\) greater than or equal to 6 for which the left-hand side is exactly \(\frac{5}{21}\) (this happens when \(n=9\)). Knowing \(n\) allows us to compute the required probability. Sufficient.
Answer: D
Mrs. Claus is selecting three reindeer from the herd for a special sleigh ride. Among the herd are Rudolph, Blitzen, and Comet. What is the probability that she chooses Rudolph, but not Blitzen and not Comet?
Assuming there are a total of \(n\) reindeer, the probability that Mrs. Claus chooses Rudolph but not Blitzen or Comet, that is, \(P(Rudolph, X, X)\) where \(X\) stands for any reindeer other than Blitzen and Comet, is:
\(3 * \frac{1}{n} * \frac{n-3}{n-1} * \frac{n-4}{n-2}\)
We multiply by 3 because Rudolph, X, X can be chosen in three different ways: {Rudolph, X, X}; {X, Rudolph, X}; and {X, X, Rudolph}.
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is \(\frac{1}{14}\).
\(P(Rudolph, \ Blitzen, \ X) = 3! * \frac{1}{n} * \frac{1}{n-1} * \frac{n-3}{n-2} = \frac{1}{14}\)
No need to solve this equation. Start substituting values of \(n\) greater than 3 and observe that as \(n\) increases, the value of the left-hand side decreases. Therefore, there is only one value of \(n\) greater than or equal to 4 for which the left-hand side is exactly \(\frac{1}{14}\) (this happens when \(n=9\)). Knowing \(n\) allows us to compute the required probability. Sufficient.
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is \(\frac{5}{21}\).
This implies:
\(P(X, X, X) = \frac{n-3}{n} * \frac{n-4}{n-1} * \frac{n-5}{n-2} = \frac{5}{21}\)
Similarly, there is no need to solve this equation. \(n\) must be at least 6, and as \(n\) increases, the left-hand side increases toward 1. Therefore, there is only one value of \(n\) greater than or equal to 6 for which the left-hand side is exactly \(\frac{5}{21}\) (this happens when \(n=9\)). Knowing \(n\) allows us to compute the required probability. Sufficient.
Answer: D
General Discussion
10 May 2024, 05:54
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Mrs. Claus is selecting three reindeer from the herd for a special sleigh ride. Among the herd are Rudolph, Blitzen, and Comet. What is the probability that she chooses Rudolph, but not Blitzen and not Comet?
It really doesn't matter who is being chosen and who is being left. For the probability we require
a) Ways 3 reindeer can be chosen from n, in which R is selected but B and C are not selected. => one seat taken by R, the remaining two seats have to be selected from n-3(R, B and C) => n-3C2
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C2}{nC3}\)
So, what you have to find is n.
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is 1/14.
Choosing two and not the third gives a probability as \(\frac{n-3C1}{nC3} = \frac{1}{14}\)
How??
a) Ways 3 reindeer can be chosen from n, in which R is not selected but B and C are selected. => one seat each taken by B and C, the remaining one seat has to be selected from n-3(R, B and C) => n-3C1
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C1}{nC3}\)
Only one variable, and you will get the answer.
Sufficient
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is 5/21.
Choosing none of the three means selecting three from remaining n-3 and gives a probability of \(\frac{n-3C3}{nC3} = \frac{5}{21}\)
Only one variable, and you will get the answer.
Sufficient
When you solve, which is not required here, n is 9.
D
It really doesn't matter who is being chosen and who is being left. For the probability we require
a) Ways 3 reindeer can be chosen from n, in which R is selected but B and C are not selected. => one seat taken by R, the remaining two seats have to be selected from n-3(R, B and C) => n-3C2
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C2}{nC3}\)
So, what you have to find is n.
(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is 1/14.
Choosing two and not the third gives a probability as \(\frac{n-3C1}{nC3} = \frac{1}{14}\)
How??
a) Ways 3 reindeer can be chosen from n, in which R is not selected but B and C are selected. => one seat each taken by B and C, the remaining one seat has to be selected from n-3(R, B and C) => n-3C1
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C1}{nC3}\)
Only one variable, and you will get the answer.
Sufficient
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is 5/21.
Choosing none of the three means selecting three from remaining n-3 and gives a probability of \(\frac{n-3C3}{nC3} = \frac{5}{21}\)
Only one variable, and you will get the answer.
Sufficient
When you solve, which is not required here, n is 9.
D
