Problem Solving

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) ¼
(C) 2/7
(D) 1/3
(E) ½


[spoiler]Source: https://www.west.net/~stewart/gmat[/spoiler]

There're only 2 sets that can be chosen from group 1 : MM (2 men) and WM (Man + woman)
P(MM) = 3C2/4C2 ( 3C2 is the no. of ways that 2 men can be picked from 3 men, 4C2 is the no. of ways that 2 persons can be picked from 4 persons - note: order is not considered, it means that WM and MW are considered as one case)
Hence, P(MM) = 3/6 = 1/2
P(WM1) = 1 -1/2 = 1/2

There're only 2 sets that can be chosen from group 2 : WW (2 women) and WM (Man + woman)
P(WW) = 2C2/3C2 = 1/3
P(WM2) = 1- 1/3 = 2/3

To form the new group - WWMM, there're 2 ways:
Combine MM of the first group with WW of the second (A)
Combine WM of the first group with WM of the second (B)
The probalilities respectively are:
P(A) = 1/2 * 1/3 = 1/6
P(B) = 1/2 * 2/3 = 1/3
P(total) = P(A) + P(B) = 1/6 + 1/3 = 1/2
Pick E.

i think answer must be 1/4. not sure of the approach i used.

OA plz..i will post my exlanation if the answer is 1/4.

lokesh r wrote:i think answer must be 1/4. not sure of the approach i used.

OA plz..i will post my exlanation if the answer is 1/4.

incorrect


The correct response is [spoiler](E)[/spoiler].

You don't need to resort to statistical formulas to handle this probability question; instead, simply tally up all the possibilities. Let {M1, M2, M3, W1} represent one group (we'll call it "group A"). Six distinct pairs are available from group A:

{M1,M2} {M1,M3} {M1,W1} {M2,M3} {M2,W1} {M3,W1}

Similarly, let {W2, W3, M4} represent the other group (we'll call it "group B"). Three distinct pairs are available from group B:

{W2,W3} {W2,M4} {W3,M4}

Any of the six pairs in group A can be selected along with any of the three pairs in group B; thus the total number of distinct groups of four riders (one pair from group A, one pair from group B) is 18. Of this total, 9 groups contain two men and two women:

{M1,M2} {W2,W3}
{M1,M3} {W2,W3}
{M2,M3} {W2,W3}
{M1,W1} {W2,M4}
{M2,W1} {W2,M4}
{M3,W1} {W2,M4}
{M1,W1} {W3,M4}
{M2,W1} {W3,M4}
{M3,W1} {W3,M4}

Accordingly, the probability of selecting two men and two women is [spoiler]9 in 18, or ½[/spoiler].



[spoiler]Courtesy: https://www.west.net/~stewart/gmat[/spoiler]

Say Group A (3M, 1W)
and group B (2W, 1M)
Total no of Ways for selecting 2 canoe riders from group A and 2 from group B= 4C2 * 4C2 = 36
No of ways of selecting canoe riders following the condition: Two canoe riders must be selected from each of two groups of campers and in that exactly 2 men and 2 women are present
The following cases arise:
1. 2Ma, 2Wb = 3
2. 2Ma, 1Wa, 1Wb =6
3. 1Ma, 1Wa, 1Mb, 1Wb= 6
4. 1Ma, 2Wb,1Mb=3
Total= 3+6+6+3=18
Probability= 1/2

brijesh wrote:Say Group A (3M, 1W)
and group B (2W, 1M)
Total no of Ways for selecting 2 canoe riders from group A and 2 from group B= 4C2 * 4C2 = 36
No of ways of selecting canoe riders following the condition: Two canoe riders must be selected from each of two groups of campers and in that exactly 2 men and 2 women are present
The following cases arise:
1. 2Ma, 2Wb = 3
2. 2Ma, 1Wa, 1Wb =6
3. 1Ma, 1Wa, 1Mb, 1Wb= 6
4. 1Ma, 2Wb,1Mb=3
Total= 3+6+6+3=18
Probability= 1/2

excuse me , i dont know how you calculate it 36
select 2 from group A: 4!/2!2!=6
select 2 from group B: 3!/2!1!=3
total=18

brijesh wrote:Say Group A (3M, 1W)
and group B (2W, 1M)
Total no of Ways for selecting 2 canoe riders from group A and 2 from group B= 4C2 * 4C2 = 36
No of ways of selecting canoe riders following the condition: Two canoe riders must be selected from each of two groups of campers and in that exactly 2 men and 2 women are present
The following cases arise:
1. 2Ma, 2Wb = 3
2. 2Ma, 1Wa, 1Wb =6
3. 1Ma, 1Wa, 1Mb, 1Wb= 6
4. 1Ma, 2Wb,1Mb=3
Total= 3+6+6+3=18
Probability= 1/2

excuse me , i dont know how you calculate it 36
select 2 from group A: 4!/2!2!=6
select 2 from group B: 3!/2!1!=3
total=18

i agree that the answer is E but i do not agree with 36
my solution is : 3/4*2/3=1/2

limestone wrote:There're only 2 sets that can be chosen from group 1 : MM (2 men) and WM (Man + woman)
P(MM) = 3C2/4C2 ( 3C2 is the no. of ways that 2 men can be picked from 3 men, 4C2 is the no. of ways that 2 persons can be picked from 4 persons - note: order is not considered, it means that WM and MW are considered as one case)
Hence, P(MM) = 3/6 = 1/2
P(WM1) = 1 -1/2 = 1/2

There're only 2 sets that can be chosen from group 2 : WW (2 women) and WM (Man + woman)
P(WW) = 2C2/3C2 = 1/3
P(WM2) = 1- 1/3 = 2/3

To form the new group - WWMM, there're 2 ways:
Combine MM of the first group with WW of the second (A)
Combine WM of the first group with WM of the second (B)
The probalilities respectively are:
P(A) = 1/2 * 1/3 = 1/6
P(B) = 1/2 * 2/3 = 1/3
P(total) = P(A) + P(B) = 1/6 + 1/3 = 1/2
Pick E.

Very Good work limestone.
Can you explain two steps for me,

what are bolded in your quote.

diebeatsthegmat wrote:

brijesh wrote:Say Group A (3M, 1W)
and group B (2W, 1M)
Total no of Ways for selecting 2 canoe riders from group A and 2 from group B= 4C2 * 4C2 = 36
No of ways of selecting canoe riders following the condition: Two canoe riders must be selected from each of two groups of campers and in that exactly 2 men and 2 women are present
The following cases arise:
1. 2Ma, 2Wb = 3
2. 2Ma, 1Wa, 1Wb =6
3. 1Ma, 1Wa, 1Mb, 1Wb= 6
4. 1Ma, 2Wb,1Mb=3
Total= 3+6+6+3=18
Probability= 1/2

excuse me , i dont know how you calculate it 36
select 2 from group A: 4!/2!2!=6
select 2 from group B: 3!/2!1!=3
total=18
i agree that the answer is E but i do not agree with 36
my solution is : 3/4*2/3=1/2

Yes U are right just i overlooked group B.

@goyalsau:

As group 1 has 3 men and 1 woman, so in which everway you pick 2 persons from 4 of them, there'll be only 2 possiblity :
2 men or 1 man with 1 woman. The total possibility is 1; if P(MM) as calculated above is 1/2, then P(MW) will be equal 1- 1/2 = 1/2. Or P(MW) -group 1 can be calculated as:
P(MW) = 3C1*1C1/4C2 ( 3C1 is no. of ways to pick 1 man from 3 men; 1C1 is no. of way to pick 1 woman from 1 woman; C is used because order is not considered)

The same for group 2:
Instead of calculating P(WM2) = 1-P(WW2) = 1 -1/3 =2/3
We can calculate: P(WM2) = 2C1*1C1/3C2 = 2/3 ( 2C1 is no. of ways to pick 1 woman from 2 women. 1C1: 1 man from 1 man. And 3C2 is no. of ways to pick 2 persons from 3)

The rule is:the sum of possibility of all cases must be equal to 1
In group 1 only 2 cases exist: MM and MW, then P(MM) + P(MW) = 1
If there were three cases - MM, WW, WM,then 1 would be equal to P(MM) + P(WW) + P(WM)