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If x and y are selected from 2, 3, 4, 5, and 6, what is the

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Director
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If x and y are selected from 2, 3, 4, 5, and 6, what is the [#permalink] New post 21 Mar 2006, 23:24
If x and y are selected from 2, 3, 4, 5, and 6, what is the probability that x*y is divisible by 4?
Director
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 [#permalink] New post 22 Mar 2006, 00:35
isn't it 2/5? :)

5C2 total possible number of picks... = 10
Favourable picks 4
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 [#permalink] New post 22 Mar 2006, 02:22
is it 1/2.

5/5C2

Thanks
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 [#permalink] New post 22 Mar 2006, 02:31
Gmate wrote:
is it 1/2.

5/5C2

Thanks


Oh yeah, 5 favourable outcomes.... 1/2
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 [#permalink] New post 22 Mar 2006, 14:57
can you guys elaborate on this one? Is there a quick way or did you guys just calclated all the outcomes? thanks
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 [#permalink] New post 22 Mar 2006, 15:36
total outcome 10
favorable outcome 5

5/10 = 1/2
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 [#permalink] New post 22 Mar 2006, 22:25
I think we get 6 favorable outcomes.

(2,4) (3,4) (4,4) (5,4) (6,4) (2,6)
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 [#permalink] New post 23 Mar 2006, 01:18
vivek123 wrote:
I think we get 6 favorable outcomes.

(2,4) (3,4) (4,4) (5,4) (6,4) (2,6)


Nope you can not count 4 and 4, because you have to choose from 2 3 4 5 and 6... there are no two 4s
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 [#permalink] New post 23 Mar 2006, 01:22
joemama142000 wrote:
can you guys elaborate on this one? Is there a quick way or did you guys just calclated all the outcomes? thanks


We calculated total outcomes by using the combination formula....
n!/(k!(n!-k!))

n! - denotes the total number of choices we have = 5
k! - denotes the number of choices we are arranging = 2

In this example n!=1*2*3*4*5
k!=1*2

And yes, the favourable outcomes are calculated by hand :) but it does not take long... max 1 minute...
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 [#permalink] New post 23 Mar 2006, 23:28
vivek123 wrote:
I think we get 6 favorable outcomes.

(2,4) (3,4) (4,4) (5,4) (6,4) (2,6)


in that case (2,2) and (6,6) would work too but it would be over 5^2 = 25

I guess Sima is right
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 [#permalink] New post 24 Mar 2006, 00:57
x*y must be a multiple a 4: 4,8,12,16,20,24,28

(x,y) sets

(2,4)
(2,6)
(3,4)
(4,5)
(4,6)

We don't have to consider (4,2) (6,2) (4,3) (5,4) amd (6,4) as we're not interested in placing but groupings.

# of ways ot pick any 2 numbers = 5C2 = 10
P = 5/10 = 1/2
Director
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 [#permalink] New post 24 Mar 2006, 01:54
I got the favourable outcomes as
2 2
2 4
2 6
3 4
4 2
4 3
4 4
4 5
4 6
5 4
6 2
6 4

Total outcomes = 25 (x can be any of the 5 nos and y can be any of the 5 nos.)

Probability = 12/25
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 [#permalink] New post 24 Mar 2006, 07:53
remgeo wrote:
I got the favourable outcomes as
2 2
2 4
2 6
3 4
4 2
4 3
4 4
4 5
4 6
5 4
6 2
6 4

Total outcomes = 25 (x can be any of the 5 nos and y can be any of the 5 nos.)

Probability = 12/25


I had the same result, because no where it is written that once you picked a number, there is one less for the next selection.

So my answer is 12/25.
  [#permalink] 24 Mar 2006, 07:53
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