M40-104

Question

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  \frac{1}{14} .
 
(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is  \frac{5}{21} .

Bunuel · Answer

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

Prateek_Sood · Answer

Bunuel  can you please help me in figuring out how can we be certain that every time we'll get the required probability as stated in 1 and 2 for some integer n.

Bunuel · Answer

Your question is not clear. If there were no value of n that satisfies the condition, the statement would be false, which is never the case in GMAT DS. The actual point is whether the value is unique, and as shown, we can be certain there is exactly one such n.

Prateek_Sood · Answer

Bunuel  Sorry my main question was whether we can be sure for the probability equations that we have set up using statement 1 and 2. Whether we’ll find some integer n ( because reindeers) that will solve that equation. It could be a non integer value as well in some cases. I can see that a value of n = 9 will solve it but fir a different set of equations and given the time constraints we can’t explore all the values

Bunuel · Answer

On GMAT DS, if a statement gives an exact probability, you can be sure there's an integer n that makes the equation true, otherwise the statement would be false, which never happens. The only question is whether that n is unique, and as shown, it is.

dhairya18 · Answer

Hi  Bunuel ,
How can you be certain that these cubic polynomial equation will yield only one relevant solution in each statement?

Bunuel · Answer

This is explained in the solution. The point is not that a cubic equation always has one relevant solution. The point is that the left hand side changes in only one direction as n increases: in (1) it decreases, and in (2) it increases. So once n = 9 gives the required value, no other relevant value of n can give the same left hand side.

arunbhati · Answer

Hi Bunuel,

Can you please explain how this is arrived

for Question:  3∗1/n∗(n−3/n−1)∗(n−4/n−2)

for 1 statement: 3* 1/n* 1/n-1* N-3/N-2

Bunuel · Answer

It is a sequential probability, without replacement.

For the main question, suppose Rudolph is chosen first. Then:

\frac{1}{n}  = probability of choosing Rudolph first.

After Rudolph is chosen, there are  n - 1  reindeer left, but Blitzen and Comet are not allowed, so there are  n - 3  acceptable choices for the second reindeer. Hence  \frac{n - 3}{n - 1} .

After that, there are  n - 2  reindeer left, and still Blitzen and Comet are not allowed, so there are  n - 4  acceptable choices. Hence  \frac{n - 4}{n - 2} .

Since Rudolph could be chosen in any of the 3 positions, we multiply by 3:

3 * \frac{1}{n} * \frac{n - 3}{n - 1} * \frac{n - 4}{n - 2}

For statement (1), we need Rudolph and Blitzen, but not Comet. Rudolph, Blitzen, and the third allowed reindeer can be arranged in  3!  ways, so:

3! * \frac{1}{n} * \frac{1}{n - 1} * \frac{n - 3}{n - 2}

M40-104

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