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Director
Joined: 17 Oct 2005
Posts: 928

If x and y are selected from 2, 3, 4, 5, and 6, what is the [#permalink]
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22 Mar 2006, 00:24
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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
Joined: 06 Feb 2006
Posts: 897

isn't it 2/5?
5C2 total possible number of picks... = 10
Favourable picks 4



Intern
Joined: 18 Jan 2006
Posts: 15

is it 1/2.
5/5C2
Thanks
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Director
Joined: 06 Feb 2006
Posts: 897

Gmate wrote: is it 1/2.
5/5C2
Thanks
Oh yeah, 5 favourable outcomes.... 1/2



Director
Joined: 17 Oct 2005
Posts: 928

can you guys elaborate on this one? Is there a quick way or did you guys just calclated all the outcomes? thanks



Senior Manager
Joined: 11 Nov 2005
Posts: 328
Location: London

total outcome 10
favorable outcome 5
5/10 = 1/2



SVP
Joined: 14 Dec 2004
Posts: 1689

I think we get 6 favorable outcomes.
(2,4) (3,4) (4,4) (5,4) (6,4) (2,6)



Director
Joined: 06 Feb 2006
Posts: 897

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



Director
Joined: 06 Feb 2006
Posts: 897

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...



Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America

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



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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
Joined: 24 Oct 2005
Posts: 659
Location: London

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



Director
Joined: 02 Mar 2006
Posts: 575
Location: France

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.










