There is a group of people consisting of 10 men and 6 women. Among the

Question

There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (manwoman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?

A) 1/30
B) 1/15
C) 7/30
D) 1/4
E) 1/2

A) 1/30
B) 1/15
C) 7/30
D) 1/4
E) 1/2

dabaobao · Accepted Answer

Veritas Prep Official Solution

P(Selecting a Married Man) = 4/10

P(Selecting the Wife of that Man) = 1/6

P(Married Couple is Selected) = (4/10)*(1/6) = 4/60 = 1/15

Explanation

We need to select a couple. Here, it doesn’t matter whether you select the man first or the woman; the two member types are different and there is only one way in which you can select the corresponding partner. You cannot select two members of the same type e.g. two men or two women. Hence we don’t need to bother with calculating the different cases of selecting the man first or the woman first.

Of course, even if we do it, we will get the correct answer. Let me show you the calculation.

The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways.

Probability of selecting a couple = 8/120 = 4/60 (same as before).

Mohammadmo · Answer

4C1/ 10C1×6C1=4/60=1/15
Option B

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kornn · Answer

Dear VeritasKarishma EMPOWERgmatRichC , After reading the article, I still do not understand when the order of picking man or woman first matters or does not matter. Here the order does not matter since picking a man is INDEPENDENT from picking a woman right? Whichever couples we pick, P(Selecting the Wife of that Man) is still 1/6. Is this the right reasoning. It is like we pick man from one pool and woman from ANOTHER pool right? And those pools are not overlapping at all. More detailed explanation will be highly appreciated

Thank you!

kornn · Answer

Dear Bunuel IanStewart , How can this question be solved by Probability Approach? I am confused whether there should 2 cases in Probability Approach: 1. Picking a man, then woman 2. Picking a woman, then man (4/6) * (1/10). Should we take those 2 cases into account? I think it is best solved by Combinatoric Approach. But I'd like to know how to solve using Probability Approach. Thank you

EMPOWERgmatRichC · Answer

Hi All,

We're told that there is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (man-woman couples). We're asked if one man and one woman are selected at random, what is the probability that a married couple gets selected. This question can be approached in a couple of different ways, but you might find it easiest to break the probability down into 2 options:

1) Select a married man first and his wife second..... (4/10)(1/6) = 4/60
2) Select a married woman first and her husband second.... (4/6)(1/10) = 4/60

Before you do anything else, it's important to note in this approach each married couple has been counted TWICE (since the same man-woman couple appears in each of the two calculations). Thus, we have to divide the overall total by 2.

Total probability of selecting a married couple = (4/60 + 4/60) / 2 = (8/60) / 2 = 4/60 = 1/15

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

IanStewart · Answer

The 'official solution' above uses what I would call a "Probability Approach". If you're picking from two separate groups, you can pick from either group first, and you'll get the right answer, no need to worry about cases. It's when you're picking from the same group (two marbles from a bag of marbles, say), and are asked about a result that could happen in different ways (getting one green and one red marble, say, which could happen red then green, or green then red) that you would need to consider cases if using a probability approach.

Mohammadmo's solution above is the fastest one to this problem, I think. There are 10*6 = 60 possible selections of a man and a woman. There are 4 married pairs we can choose. So the answer is 4/60 = 1/15.

KarishmaB · Answer

Yes, the question gives you that one man and one woman are selected. So there are two distinct pools from which you have to select one person each. Hence there is no order of selection. It is like if you have to create a meal using one main and one side dish, it doesn't matter which you pick first - main or side.

If instead, we were to find the ways of selecting a man and a woman out of 16 people, there we can select a man and then a woman or a woman and then a man (as further cases, we could select a man and a man or a woman and a woman)

Louis14 · Answer

Here's one way to approach this rather interesting question.

So, we are told that there are 4 couples among a total of 16 people. This means that out of 16 people, 8 are married. Okay. So, 8/16*1/15 = 1/2*1/15= 1/30. (Basically, here, I first found out the probability of selecting a married person i.e. 8/16, and then multiplied that with 1/15, which is the probability of finding the spouse of that married person). Let's say that in the above calculation we got a "married man" in the first place, and his spouse in the second place.

Now, what if we selected the spouse first, and then the "married man"? That's also a possibility. So, we would repeat the exact same calculation as above: 8/16*1/15 = 1/2*1/15= 1/30. [This is similar to asking that how many ways are there to arrange a Husband and Wife? 2! = 2].

Why did we do it twice? Because the question demands us to find the probability that a married couple gets selected. So, what we need to find is basically this:

P(finding a married man AND his spouse) or P(finding a married woman AND her spouse)= 1/30 + 1/30 = 2/30 = 1/15

Alternative method:

8/16*1/15*2! = 1/2*1/15*2= 1/15

Hope it helps!

wishmasterdj · Answer

Since a man and a woman being chosen are given in the question, they are 2 separate events:
Thus, we can choose any of the 4 married men from the 10 men, but for whoever we choose there is only 1 wife amongst the 6 women.

4/10 * 1/6 = 1/15

gmatpro15 · Answer

we can choose any of the 4 married men from the 10 men, but for whoever we choose there is only 1 wife among the 6 women.

4/10 * 1/6 = 1/15

HarishD · Answer

10-M
6-W (4 married couple)

(M1W1) (M2W2) (M3W3) (M4W4) M5 M6 M7 M8 M9 M10 W5 W6

Number of way 1 men and 1 women are selected = 10C1 * 6C1 = 60 groups

Number of  arrangements  if selected M and W are married = 8*1 (either men or women are selected then his/her partner)
Number of  groups  for above arrangements (it can be arranged in 2! ways) = 8*1/2! = 4 groups

P = 4/60

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