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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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!
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There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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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 :)
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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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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

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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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varotkorn wrote:
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?


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.
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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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varotkorn wrote:
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!


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)
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There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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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! ;)
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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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
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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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
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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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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Re: There is a group of people consisting of 10 men and 6 women. Among the [#permalink]
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