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Probability Surfeit

Expert replies
by dtweah » Sat May 09, 2009 10:11 am
Three men and three women are assigned different numbers selected at random from the integers 1 through 6. What is the probability the three numbers assigned to the men are all greater than the three numbers assigned to the women?

(a) 1/6
(b) 1/9
(c) 1/12
(d) 1/18
(e) 1/20
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Source: — Problem Solving |

by DeepakR » Sat May 09, 2009 10:41 am
3 numbers assigned to Men are all greater than that of women. So out of 1 to 6:

The 3 Men can take the following 3 numbers
M - 4,5,6 and women would be W - 1,2,3

So probability of getting 4,5 and 6 = 1/3*1/2*1=1/6 = A.)

-Deepak
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by dtweah » Sat May 09, 2009 1:31 pm
DeepakR wrote:3 numbers assigned to Men are all greater than that of women. So out of 1 to 6:

The 3 Men can take the following 3 numbers
M - 4,5,6 and women would be W - 1,2,3

So probability of getting 4,5 and 6 = 1/3*1/2*1=1/6 = A.)

-Deepak
Take another shot Deepak.
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by artistocrat » Sat May 09, 2009 3:09 pm
1. The men take spots 1,2,3 and the women take spots 4,5,6.

2.The number of ways of organizing the men in a group of 3 is 6, and same with women is 6.

3. So, the total number of ways of organizing the men and women using multiplication rule is 6*6=36.

4. The total number of ways to organize 6 people is 6*5*4*3*2*1=720.

5. So 36/720=1/20
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by dtweah » Sat May 09, 2009 4:02 pm
artistocrat wrote:1. The men take spots 1,2,3 and the women take spots 4,5,6.

2.The number of ways of organizing the men in a group of 3 is 6, and same with women is 6.

3. So, the total number of ways of organizing the men and women using multiplication rule is 6*6=36.

4. The total number of ways to organize 6 people is 6*5*4*3*2*1=720.

5. So 36/720=1/20
Good Work Aris! You solved it as in the OA.

Alternatively let 456 be 3 red balls and the rest three black balls

First red 3/6

second red 2/5
3rd red 1/4

Prob men pick all red balls on three random draws is 3/6 x 2/5 x 1/4=1/20
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by avenus » Mon May 11, 2009 4:10 am

Hey,
try it without the restriction of different numbers, i.e, that each man/woman gets one random integer n, 1<=n<= 6 (obtained by rolling a dice, for example). This makes it more complicated. Who wants to give it a go??
What would then the result be??

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