Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.

I have trouble with this question, help anyone?

So for 1), we treat each couple as one unit, and we need to pick 2 out of this six units. We get 6C2, and 12C4 means the number of combinations we can get from 12 people. Ok, this one I get it.

For 2), what is the logic to use 12C1 x 10C1 x 8C1 x 6C1 and 12C4 x 4!?

It is tough for me too. I agree with the answers given by lovely_baby.

(a) 6C2/12C4 = 1/33
(b)The number of ways the chosen 4 can be unmarried = 12C1 * 10 C1 * 8C1 * 6C1 = 12*10*8*6

Because the above number is a permutation (order is important), we need to convert it into combination (order is not important as we are choosing) i.e., The actual ways without regard for order = (12*10*8*6)/4!
= 5*8*6
The required probability = (5*8*6)/12C4 = 16/33

(c) The probability that exactly one is married couple = 1-(a)-(b) = 16/33

Can anybody solve (c) in the way we have solved (a) & (b)? Somehow, I am in the loop and unable to come out of it.


_________________

Awaiting response,

Thnx & Rgds,
Chandra

Thanks for the answer. But lovely_baby,

We need to divide by 3! because the number of ways with 3 items (1 couple and 2 members who are not a couple) are calculated here and order is not important here. infact I too got the answer with 2! division but could not justify it.


_________________

Awaiting response,

Thnx & Rgds,
Chandra

I see it like this:

c) exactly one married couple is among the 4.

A A, B B, C C, .., E E
twelve people; six couples

Favorable outcome (example):
A A B C

6C1 ways to select one couple (A A)

10C2 ways to select the remining two people (B C) - but this includes selecting another couple (like B B, or C C); so subtract 5C1

Favorable combinations = 6C1*(10C2 - 5C1) = 240
Total combinations = 12C4 = 495
P(AABC) = 240/490 = 16/33

Excellent!! I got the stuff very clearly. I was getting the answer from set theory but not in a direct way.

Thanks d-dogg!!


_________________

Awaiting response,

Thnx & Rgds,
Chandra

Christoph,

It is tough for me too. I agree with the answers given by lovely_baby.

(a) 6C2/12C4 = 1/33
(b) While choosing 4, no couple shoule be present. Let us pick 1 out of 12. In the remaining 12, the first person's spouse is present and thus has to be skiiped because no couple should be present in 4. So, we can choose 2nd person out of 10. This 2nd person's spouse is present in the remaining 9. So, the 3rd person has to be from the remaining 8. Similarly, 4th person has to be from the remaining 6.


The total number of ways the chosen 4 can form no couples = 12*10*8*6
Here, all the number of ways are accounted i.e. ABCD, ABDC, ADCB etc. But, when we pick 4 persons at random, they can be in any order. All of ABCD, ABDC, ADCB etc are one and the same as for as the group is concerned. Thus, we need to divide the (12*10*8*6) with 4!

In other words,
Because the above number is a permutation (order is important), we need to convert it into combination (order is not important as we are choosing) i.e., The actual ways without regard for order = (12*10*8*6)/4!
= 5*8*6
The required probability = (5*8*6)/12C4 = 16/33

(c) If you observe care fully, 1 = P(a) + P(b) + P(c) from the set theory because Total probability (i.e., 1) = Probability that two married couples are chosen + Probability that one married couple is chosen + Probability that NO married couple is chosen

The probability that exactly one is married couple = 1-P(a)-P(b) = 16/33

The above method is easy by observation. However, if P(c) is asked straight, then you may work out as follows.

Let P(c) = Probability that exactly one is married

The number of ways of choosing 4 such that only 1 is married couple = 6C1 * 10C2 - 6C1*5C1 = 6*45 - 6 * 5 = 240

This is very well explained d-dogg

"6C1 ways to select one couple (A A)

10C2 ways to select the remining two people (B C) - but this includes selecting another couple (like B B, or C C); so subtract 5C1

Favorable combinations = 6C1*(10C2 - 5C1) = 240 "

Total space = 12C4

P(c) = 240/12C4 = 16/33

The second method is easier if you understand the logic in it. It is faster too. If you are uncomfortable with it, go for the first method where you should be good at breaking up the unknown into known problems.

neelesh wrote:
This question is probably a simple one for most people on this forum here, but wanted to see what approach to use for such questions. Somehow my appeoach is not matching with the OA.

Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.



_________________

Awaiting response,

Thnx & Rgds,
Chandra

thx honguhu and mallelac !

how do we know the number of permutations respectively 4! in B) and 2! in A) ?

Don't forget the difference bt combinations and permutations, anirban16.