Artificial intelligence computer HAL 9000 randomly picks three

Question

Artificial intelligence computer HAL 9000 randomly picks three distinct integers from between 1 and 9000 inclusive. If the first number picked is a, the second number picked is b, and the third number picked is c, what is the probability that a > b > c ?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

Bunuel · Accepted Answer

Official Solution:

Artificial intelligence computer HAL 9000 randomly picks three distinct integers from between 1 and 9000 inclusive. If the first number picked is a, the second number picked is b, and the third number picked is c, what is the probability that a > b > c ?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

A.  \frac{1}{60} 
B.  \frac{1}{30} 
C.  \frac{1}{20} 
D.  \frac{1}{6} 
E.  \frac{1}{3}

The question basically asks about the probability that three numbers picked are in decreasing order ( a > b > c ).
 
Any three numbers can be picked in  3! = 6  ways, and only one sequence from these 6 will be in decreasing order. So, the probability is  \frac{1}{6} .
 
For example, 2, 5, and 6 can be drawn in 6 ways ({firs number, second number, third number}): {2, 5, 6}, {2, 6, 5}, {5, 2, 6}, {5, 6, 2}, {6, 2, 5} and {6, 5, 2}. Only one from these 6 is in decreasing order: {6, 5, 2}.

Answer: D

chetan2u · Answer

Do not worry about how many numbers are to be chosen from. What is important is how many do we have to choose, because when they are rearranged only 1 out of the different ways they can be rearranged will have a>b>c.

So ways to arrange the chosen 3 = 3*2*1=6 ways
Only 1 out of these 6 ways will have the numbers in descending order, so probability =  \frac{1}{6}

D

Kunni · Answer

Hi,

I did not understand the answer, could you please elaborate it.

Regards
Kunak

chetan2u · Answer

Hi,

It really doesnt matter how many number sare in the group..

Take this question itself...
Ways to pick three numbers when order matters = 9000*8999*8888
Now ways to pick when the three number are in a specific order => same as picking 3 numbers without any order => 9000C3=9000*8999*8998/3!

Probability =  \frac{9000C3}{9000*8999*8998}=\frac{9000*8999*8998}{3!*9000*8999*8998}=\frac{1}{3!}

ScottTargetTestPrep · Answer

There are 9000C3 ways to pick 3 numbers (in any order) from 9000. Any one of these 9000C3 sets of 3 numbers can occur equally likely as the other. However, once a set of 3 numbers is picked, there is only a 1/6 chance that the numbers picked are in descending order. For example, if 1, 10 and 100 are picked, there are 6 ways to order them: (a, b, c) = (1, 10, 100), (1, 100, 10), (10, 1, 100), (10, 100, 1), (100, 1, 10) and (100, 10, 1). We see that only the last one is in descending order, i.e., a > b > c. Therefore, the probability that a > b > c is 1/6.

Answer: D

tommychudw · Answer

Keep it simple.

Consider {1, 2, 3}.

There is only 1 way to pick where a > b > c:
Pick 3, then 2, and then 1.

1/3 * 1/2 * 1/1 = 1/6

Answer (D).

Artificial intelligence computer HAL 9000 randomly picks three

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Artificial intelligence computer HAL 9000 randomly picks three

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards