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Could you elaborate why the numerator is 3C1 * 1 ?
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Moritz279
Could you elaborate why the numerator is 3C1 * 1 ?

Total Six Men are there in that three are married. We need to select one from the three men so it is 3C1.
Also we need select his wife that is 1(we can't use 3c1 for selecting women since we can't create a new couple)

So the possible way is 3C1*1.

One more method is 3/6*1/7=1/14

Posted from my mobile device
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Moritz279
Could you elaborate why the numerator is 3C1 * 1 ?

Hi Moritz279,

As we have 3 couples in the group,

The probability of Selecting Husband out of the 3 married couples is \(C^3_1\). --- (1)

Now any couple will have a husband and his wife. Now from (1), when I select a husband, the probability of selecting his wife will automatically become 1 as their will be no other combinations with that husband (No pun intended ;) )

Hence, The probability of Selecting Wife of the selected husband is \(1\).

Hope it was helpful :)
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iamsiddharthkapoor
A dance group consists of 6 men and 7 women. There are 3 married couples in the group. If 2 people have to be selected – 1 man and 1 woman, what is the probability that one of the couples will be selected?

A. 1/14

B.1/7

C. 1/6

D. 1/3

E. 1/2
Required Probability = No of ways couples can be selected / total no of ways in which 1 man and 1 woman can be selected
= 3 / (6 x 7) = 1/14

A is correct.

Study from the best self learning GMAT quant course:
https://www.udemy.com/course/best-gmat- ... 37602183BB
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The number of ways for selecting a husband out of 3 is \(^3{C_1}\)

The number of ways for selecting the wife of an already selected husband: 1

Total ways: \(^6{C_1} * ^7{C_1}\)

=> \(\frac{[^3{C_1}]}{ [^6{C_1} * ^7{C_1}]}\)

=> \(\frac{3 }{ 42} \)

=> \(\frac{1}{14}\)

Answer A
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MathRevolution
The number of ways for selecting a husband out of 3 is \(^3{C_1}\)

The number of ways for selecting the wife of an already selected husband: 1

Total ways: \(^6{C_1} * ^7{C_1}\)

=> \(\frac{[^3{C_1}]}{ [^6{C_1} * ^7{C_1}]}\)

=> \(\frac{3 }{ 42} \)

=> \(\frac{1}{14}\)

Answer A

But with this in mind, why dont we consider the possibility of selecting the wife BEFORE the husband? wouldn't that be 1/14 * 2 (for husband->wife and for wife -> husband)?
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siddharthkapoor
A dance group consists of 6 men and 7 women. There are 3 married couples in the group. If 2 people have to be selected – 1 man and 1 woman, what is the probability that one of the couples will be selected?

A. 1/14

B.1/7

C. 1/6

D. 1/3

E. 1/2

Can someone tell.

Why this selection will be wrong?

selecting 2 from 6 people(married couple)

=> 6c1*1c1/ 6c1*7c1 = 1/7
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MathRevolution
The number of ways for selecting a husband out of 3 is \(^3{C_1}\)

The number of ways for selecting the wife of an already selected husband: 1

Total ways: \(^6{C_1} * ^7{C_1}\)

=> \(\frac{[^3{C_1}]}{ [^6{C_1} * ^7{C_1}]}\)

=> \(\frac{3 }{ 42} \)

=> \(\frac{1}{14}\)

Answer A

Although, i got this question correct but i have a doubt.

i was trying to resolve in a different way, but not able to found error in it.

there are 6 people in 3 couple.... so we can select one in 6c1 ways and second in 1 way.

So p= 6/6c1*7c1

thanks
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Married Man can be selected in 3C1 ways out of 6C1 possibility. Woman can be selected in 1 ways out of 7C1 possibility
3C1*1
____________
6C1*7C1

VeritasKarishma chetan2u why not max. possibility of selection is 13C2?
i.e 3C1*1
------------ ways
13C2
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Nirmesh83
Married Man can be selected in 3C1 ways out of 6C1 possibility. Woman can be selected in 1 ways out of 7C1 possibility
3C1*1
____________
6C1*7C1

VeritasKarishma chetan2u why not max. possibility of selection is 13C2?
i.e 3C1*1
------------ ways
13C2


13C2 would include possibilities of choosing male in both selection or females in both selection, a situation that would not be correct as per quest.
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these kind of question should specify if we are picking from the entire group or from each group. because picking from the entire group leads to a very different answer.

if we pick from the entire group we are considering also the possibility of picking 2 man or 2 women. thus the total number of ways to pick is 13C2. the number of subgroups in the 13C2 that contain a man and the relative wife are 3...

but i guess that with "1 man and 1 woman" you wanted to tell us that we are picking 1 man from the 6 man and 1 woman from the 7 woman...
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siddharthkapoor
A dance group consists of 6 men and 7 women. There are 3 married couples in the group. If 2 people have to be selected – 1 man and 1 woman, what is the probability that one of the couples will be selected?

A. 1/14

B.1/7

C. 1/6

D. 1/3

E. 1/2
Total possible outcomes when picking one men and one women = 6*7 = 42

Number of possible outcomes when picking married couples = 3

Probability = Possitive outcomes / total outcomes = 3/42 = 1/14 -> A
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From 6 men 3 are married.
So selecting married one is 1/2
From 7 women 3 are married, so selecting married is 3/7
Probability of picked person being pair is 1/3 as 1 of 3 married people in either group will be a pair of married person in another group.
So we multiply:
1/2*3/7*1/3=1/14
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