What is the probability that a man and woman selected at random amongs

Question

What is the probability that a man and woman selected at random amongst 4 married couples happen to be married to each other?

A. 1/4
B. 1/8
C. 1/16
D. 1/20
E. 1/30

Juliadorvard · Answer

First we choose a men, all the options will be correct, so probability is 4/4, then we the probability to choose his wife is 1/4 (as there is only 1 wife out of 4 women).
4/4*1/4=1/4
Answer A

whatsarc · Answer

A: 1/4

4 married couples -----> 4 men and 4 women

Probability = favorable outcomes/ total outcomes

Total outcomes =  one man out of 4 x one woman out of 4 = 4C1 x 4C1 = 16

Favorable outcomes =  4 (total pairs possible) out of all 16

P= 4/16 = 1/4

Vibhatu · Answer

probability of selecting 1st person= 1 (Men or Women)
probability of selecting 2nd person= 1/4 (Men or Women)

Probability= 1*1/4=1/4

Purnank · Answer

I think simple way it is like 1/4 
one couple out of 4 couples.

Is my approach correct?

sameerssit · Answer

Probability of selecting 1 man= 4/8= 1/2

AND(MULTIPLICATION FOR AND)

Probability of selecting 1 Woman= 4/8= 1/2

P(1 M and 1 W)= 1/2 x 1/2 = 1/4

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SergejK · Answer

Bunuel  But why can't we calculate 2/8 x 1/7=1/28? As we are selecting 2 out of 8, we know this must be the couple. Or is it because we are only accounting for one specific couple and disregarding all other couples if we count like this?

To be honest, the first thing I came up with was to take 4/4 for a man and 1/4 for a women, but I counted it twice, as I could either choose the woman first or the man. But then I will get to 1/2, right? 4/4x1/4 + 1/4x4/4= 1/2. Why don't I have to account for those 2 possibilities?

SergejK · Answer

EDIT: For those who might have the same problem as I did, coming back to this question I see why I failed back then. The question is very unspecific: What is the probability that A man and A woman (meaning any couple) happen to be married to each other. We actually don't care which one, just a couple out of those eight. My earlier approach was focusing on specific 2 people, however, couples could have been constructed from any of the other couples so zeroing in only on 2 specific subjects out of 8 is not accounting for all possibilities, as other couples could also be formed that are married to each other. So my approach of 2/8 x 1/7=1/28 was wrong as there is no sequence, we are randomly choosing this unspecified one couple. If I have to calculate it like that, to account for search for the unspecified couple, we just choose any of the 8 and then look who is left that could ACTUALLY match. So instead of 2/8 x 1/7 we count 8/8 x 1/4 because any person could be selected first and then only 4 of the left could match, either man, if we first selected a woman or a woman if we first selected a man. We are not asked what is the probability that AFTER selecting a man, his wife is selected or vice versa. If the order of selection is not specified and no sequence is provided, we can be as unspecific with our calculation as possible. That is why we don't have to add the probability that a woman is selected first and man second to the probability that a man is selected first and a woman second, like I proposed in my second question. Order is not specified, hence, we are not considering any order preferences at all. Also, adding both probabilities would only be possible if we had an either or situation, a situation with mutually exclusive events. In our case, the probability of selecting married couple 1 or 2. As we can't select different couples in the same selection we would add those probabilities together to get 1/4+ 1/4=1/2. However, selecting man or a woman first is NOT mutually exclusive as they lead to the same event: bringing together the same married couple.

However, the easiest way to solve such a problem is to count all possible cases and define what are the sought cases. Here, it is 4x4=16 and 4 married couples. 4/16=1/4.

However, the easiest way to solve such a problem is to count all possible cases and define what are the sought cases. Here, it is 4x4=16 and 4 married couples. 4/16=1/4.

raj-boro · Answer

the ANSWER is 1/4. total cases= 4c1 x 4c1. One man out of 4 and One woman out of 4. Favorable cases= 4c1. One couple out of four. P= 4c1/ 4c1 x 4c1=1/4 However, if any two persons were chosen, the total cases would have been 8c2. Happy learning.

luisdicampo · Answer

Deconstructing the Question  
There are 4 married couples → 4 men and 4 women.
We randomly select 1 man and 1 woman.
We want the probability that they are married to each other.

Step-by-step  
Total possible man–woman pairs:
 4 \cdot 4 = 16

Each man has exactly 1 wife, so there are:
 4 
married pairs.

Thus,
 P=\frac{4}{16}=\frac{1}{4}

Answer:     (A) 1/4

What is the probability that a man and woman selected at random amongs

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What is the probability that a man and woman selected at random amongs

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