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).

arushi118 · Answer

Total ways to pick 3 numbers from 1 to 9000 -> 9000c3*(3!)
When you pick 3 numbers, there will only be 1 way in which a>b>c -> So number of ways to pick 3 numbers and a>b>c -> 9000c3*1

So probability = (9000c3*1)/(9000c3*6) = 1/6
D

Artificial intelligence computer HAL 9000 randomly picks three

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Artificial intelligence computer HAL 9000 randomly picks three

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