Probability Question

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mindruna: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Mon Mar 19, 2007 6:23 pm; Location: Cambridge, MA

Probability Question

by mindruna » Fri Aug 10, 2007 5:14 am

I still struggle with this type of question. Can anyone help? Thanks.

Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

A) 1/6
B) 1/4
C) 2/7
D) 1/3
E) 1/2

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Source: Beat The GMAT — Problem Solving | Fri Aug 10, 2007 5:14 am

ratindasgupta: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Jun 11, 2007 10:32 pm; Location: Mumbai

Re: Probability Question

by ratindasgupta » Fri Aug 10, 2007 6:19 am

mindruna wrote:I still struggle with this type of question. Can anyone help? Thanks.

Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

A) 1/6
B) 1/4
C) 2/7
D) 1/3
E) 1/2

I think the answer is E) 1/2

If 2M and 2W have to be selected keeping in mind 2 from each group then these are the following options
a) 2M from 1st group and 2W from 2nd group or
b) 1M & 1W from 1st group and 1M & 1W from the 2nd group or
c) 1W & 1M from the 1st group and 1W & 1M from the 2nd group or
d) 1M & 1W from 1st group and 1W & 1M from the 2nd group or
e) 1W & 1M from the 1st group and 1M & 1W from the 2nd group.
Whew!

so
a) 1/6
b) 1/12
c) 1/12
d) 1/12
e) 1/12

a + b + c + d + e = 1/2.

Wot's the OA?

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mindruna: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Mon Mar 19, 2007 6:23 pm; Location: Cambridge, MA

by mindruna » Fri Aug 10, 2007 6:31 am

Thanks Ratindasgupta!

1/2 is the OA. How are you getting 1/6 for a and 1/12 for the rest?

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ratindasgupta: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Jun 11, 2007 10:32 pm; Location: Mumbai

by ratindasgupta » Fri Aug 10, 2007 6:50 am

mindruna wrote:Thanks Ratindasgupta!

1/2 is the OA. How are you getting 1/6 for a and 1/12 for the rest?

i was trying to avoid the gory details. hee hee. anyway here goes.

basically its a dependant event problem

if u need 2m from 1st group then its 3/4 x 2/3 ((Probability of selecting 1st man is 3/4 and 2nd man is 2/3)

and 2w from the 2nd group is 2/3 x 1/2 (Probability of selecting 1st woman is 2/3 and 2nd woman is 1/2)

so 3/4 x 2/3 x 2/3 x 1/2 = 1/6

and so on and so forth. Just list the probabilities down.

1st Group 2nd Group
a) 2M 2W 3/4 x 2/3 x 2/3 x 1/2 = 1/6
b) 1M & 1W 1M & 1W 3/4 x 1/3 x 1/3 x 1 = 1/12
c) 1W & 1M 1W & 1M 1/4 x 1 x 2/3 x 1/2 = 1/12
d) 1M & 1W 1W & 1M 3/4 x 1/3 x 2/3 / 1/2 = 1/12
e) 1W & 1M 1M & 1W 1/4 x 1 x 1/3 x 1 = 1/12

Once this is done, add all figures.

I'm sorry i dont have an easier way to solve this. I doubt there is one. Where is this problem from btw?