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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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probability

by sidon » Fri Apr 09, 2010 11:36 am
Hi ,

I am not sure I understand this one :

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 was thinking D 1/3
my calc was :
to be able ro have 2w and 2m I the following must B
1. group A : 1w and 1m group B 1w and 1 m
2. group A : 2m group B 2w
no other way to have 2w and 2m

so for option one
1/4*1/3 = 1/6
for option tow
1/3 * 1/3

seems to be wrong - right answeer is 1/2 ...

can some on ehelp ?


Thanks,
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by papgust » Fri Apr 09, 2010 7:04 pm
G1: 3M, 1W (4 people)
G2: 1M, 2W (3 people)

We need to select 2 people from each group

So, Total combination is 4C2 * 3C2 = 6*3 = 18

Consider 2 scenarios where you can select 2 Men and 2 Women,

1. 2 Men from G1 AND 2 Women from G2
OR
2. 1 Man & 1 Woman from G1 AND 1 Man & 1 Woman from G2


Let's calculate now,

1. 3C2 * 2C2 = 3
OR
2. 3C1 * 1C1 * 1C1 * 2C1 = 3*2 = 6

(1) + (2) = 3+6 = 9

Probability (two men and two women will be selected) = 9/18 = 1/2
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by sidon » Sat Apr 10, 2010 10:09 am
Hi ,

Thanks for answering .
I am not sure I understand the second part of the explanation - how you calculated the probablity of the groups.

I was also hoping for some one to explain why my calc is wrong .
Can anyone help ?

Thanks,
Sidon
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by thephoenix » Sun Apr 11, 2010 4:37 am
probability=favorable case / total cases

total case=4C2*3C2{2 from each group }=18

favorable cases=from group 1 1men 1 women and from group 2 1 men and 1 women=3*1*1*2=6
or
from group 1 2 men and from group 2 2 women=3*1=3
total favorable cases=9
P=9/18=1/2
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