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First team = man1, man2, man3, woman [ M1, M2, M3, W ]
Possible pick of 2 = [ M1 & M2 ], [ M1 & M3 ], [ M2 & M3 ], [ M1 & W ], [ M2 & W ] and [ M3 & W ]
So.... total scenario = 6 ...... scenario of 2 men = 3 .....so probability of 2 men = 3 / 6 = 1 / 2 .......
Scenario of 1 woman & 1 man = 3 ...... so probability of 1 woman & 1 man = 3 / 6 = 1 / 2 ......

Second team = woman1, woman2, man [ W1, W2, M ]
Possible pick of 2 = [ W1 & M ], [ W2 & M ], [ W1 & W2 ]
So... total scenario = 3 ... scenario of 1 man & 1 woman = 2 ....so probability of 1 man & 1 woman = 2 / 3 .....
Scenario of 2 women = 1 ...... so probability of 2 women = 1 / 3 .......

Now....2 man & 2 woman can be taken in 2 ways......

2 W from second team and 2 M from first team...probability of dis scenario = [ 1 / 3 ] × [ 1 / 2 ] = 1 / 6

1 M & 1 W from 2nd team and 1 M & 1 W from 1st team...probability of dis scenario = [ 2 / 3 ] × [ 1 / 2 ] = 2 / 6

So....all probability = [ 1 / 6 ] + [ 2 / 6 ] = 3 / 6 = 1 / 2

! nah id win!
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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?

First group: 3M + 1W
Number of ways to select 2 canoe riders = 4C2 = 6

Second group: 2W + 1M
Number of ways to select 2 canoe riders = 3C2 = 3

Total number of ways to select 2 canoe riders = 6*3 = 18

Ways to select 2M + 2W: -
1M + 1W from first group & 1W + 1M from second group; Number of ways = 3C1*1 *2C1*1 = 3*2= 6
2M from first group and 2W from second group; Number of ways = 3C2*2C2 = 3
Total number of ways to select 2M + 2W = 6+3 = 9

The probability that two men and two women will be selected = 9/18 = 1/2

IMO E
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there are 2 ways in which 2M and 2W can be chosen from the 2 given groups:

case 1:
P(MM) from G1= 3/4 * 2/3 * 2!/2! = 1/2
P(WW) from G2= 2/3 * 1/2 *2!/2! = 1/3
P(MM) and P(WW)= 1/2 * 1/3 = 1/6

case 2:
P(MW) from G1 = 3/4 * 1/3 * 2! = 1/2
P(MW) from G2 = 1/3 * 2/2 * 2! = 2/3
P(MW) and P(MW) = 1/2 * 2/3 = 1/3

So total probability can be calculated by combining the probabilities from the above two cases:
case 1 or case 2 = 1/6 + 1/3 = 9/18 = 1/2
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Group 1: 3M, 1W
Group 2: 1M, 2W

Two riders to be chosen from each group.

To get two men and two women, what are the possible cases?

(1) 2 Men from Group 1 AND 2 Women from Group 2.

3C2 x 2C2 = 3.

(2) 1 Man and 1 Woman from Group 1 AND 1 Man and 1 Woman from Group 2

3C1 x 1C1 x 1C1 x 2C1 = 3 x 1 x 1 x 2 = 6

No other cases are possible.

Total Outcomes = Number of ways of choosing 2 from Group 1 AND 2 from Group 2 = 4C2 x 3C2 = 6 x 3 = 18.

Our required Probability = (3 + 6) / 18 = 9/18 = 1/2. Choice E.

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