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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
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

If x and y are selected from 2, 3, 4, 5, and 6, what is the [#permalink]  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
Joined: 05 Feb 2006
Posts: 902
Followers: 1

Kudos [?]: 49 [0], given: 0

isn't it 2/5?

5C2 total possible number of picks... = 10
Favourable picks 4
Intern
Joined: 18 Jan 2006
Posts: 15
Followers: 0

Kudos [?]: 0 [0], given: 0

is it 1/2.

5/5C2

Thanks
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Director
Joined: 05 Feb 2006
Posts: 902
Followers: 1

Kudos [?]: 49 [0], given: 0

Gmate wrote:
is it 1/2.

5/5C2

Thanks

Oh yeah, 5 favourable outcomes.... 1/2
Director
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

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: 332
Location: London
Followers: 1

Kudos [?]: 8 [0], given: 0

total outcome 10
favorable outcome 5

5/10 = 1/2
SVP
Joined: 14 Dec 2004
Posts: 1707
Followers: 1

Kudos [?]: 50 [0], given: 0

I think we get 6 favorable outcomes.

(2,4) (3,4) (4,4) (5,4) (6,4) (2,6)
Director
Joined: 05 Feb 2006
Posts: 902
Followers: 1

Kudos [?]: 49 [0], given: 0

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: 05 Feb 2006
Posts: 902
Followers: 1

Kudos [?]: 49 [0], given: 0

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
Followers: 1

Kudos [?]: 6 [0], given: 0

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: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

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: 660
Location: London
Followers: 1

Kudos [?]: 7 [0], given: 0

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: 582
Location: France
Followers: 1

Kudos [?]: 34 [0], given: 0

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.

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