Probability Problem

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Probability Problem

by gmatNooB8787 » Thu May 10, 2012 2:59 pm
Problem : At a motivational forum, a group of 1,500 people are separated into two groups, the Leaders and the Followers. 1,000 people are assigned to the Followers and 500 are assigned to the Leaders. Among these people, there are 75 married couples, each consisting of 1 Leader and 1 Follower. If 1 person is chosen randomly from each group, what is the probability that the two people chosen will be a married couple?

{
Is this different from asking : If 1 person is chosen randomly from each group, what are the different ways for choosing such that the two people chosen will be a married couple?

Would the ans be different.
becoz what i initially thought was to calculate individually each way of choosing one spouse and then the other. It gave me the double , since there are two ways , i.e. either u choose the husband or the wife first.

i.e. 75/1,000 × 1/500 + 1/1,000 × 75/500 = 3 / 10000

Please clarify.
}

A. 3/20000
B. 1/4500
C. 19/1800
D. 1/75
E. 1/20



Soln: The solution is that 75/1,000 × 1/500 = 15/1,000 × 1/100 = 15/100,000 = 3/20,000.

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by Thiagaraj » Fri May 11, 2012 9:22 pm

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by aneesh.kg » Fri May 11, 2012 11:11 pm
I understand your problem and will only address your doubt in this post.

Say the 75 married couples are: (M1,W1), (M2,W2)... (M75,W75).

When you have to select 'A Husband 1st & Wife 2nd',
say M3 is selected first and then W3 is selected. Great! We have a couple (M3, W3).

Now, when you select 'A Wife 1st & Husband 2nd',
say W3 is selected first and then M3 is selected. Oh! We have the couple (W3, M3) again! Damn!

Do you see that this happening to each and every couple?
Every couple, which should've been counted just once is being counted twice. If you have a (M17,W17), you have a (W17, M17) as well. If you have a (M41, W41), you have a (W41, M41) as well.
No wonder your answer is twice of the required answer.

If this makes sense then Congratulations on getting a very important concept clarified.
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by gmatNooB8787 » Sun May 13, 2012 4:15 am
Thanks a lot aneesh,
Then i guess the question "what are the different ways for choosing "
and "what is the probability that the two people chosen will be a married couple" are different.

Because in the first case (M1,W1) and (W1,M1) are different as in 2 different ways.
In the second case both are same. right ?

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by aneesh.kg » Sun May 13, 2012 5:57 am
No. No.

Phrases like "what are the different ways of choosing" or "what are the ways of choosing different couples" will never be used because such a phrase will be an oxymoron. If the word 'chosen' is mentioned in the problem, it signifies that only the members in the teams are important. Their orders are NOT important. The word 'different' will contradict the word 'chosen'.

Three Simple Rules of Permutations & Combinations:
(i) Words like "Choosing"/"Selecting"/"Picking" denote Combinations (nCr)
(ii) "Arrangement"/"Distribution" of n objects on n places is given by n!
(iii) 'AND' is 'multiplication', 'OR' is 'Addition'
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by Jeevanantham » Wed May 16, 2012 12:47 pm
Hi All,

Please explain this problem little bit more, I don't understand the concept.

Thanks
Jeeva.

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by aneesh.kg » Wed May 16, 2012 9:09 pm
Hi Jeevanantham,

Before I solve the problem in detail, let's look at the definition of Probabilty:
Probability = (No. of Favourable outcomes)/(Total No. of Outcomes)

The denominator (Total No. of outcomes) is based on how the selection is being made in the problem. The Problem says that one person is selected from the Leaders and one from the Followers.

Total No. of Outcomes = (No. of ways of selecting a follower) AND (No. of ways of selecting a Leader) = 1000C1*500C1 = 1000*500

The numerator (No. of favourable outcomes) are those outcomes in which both are people chosen are married, or those outcomes in which a couple is chosen from the available couples.

No. of Favourable Outcomes = Select one couple from 75 couples = 75C1 = 75

Required Probability = 75/(1000*500) = 3/20000
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