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abhudayaharp

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24 Jul 2025, 19:22

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I like the solution - it’s helpful.

Unknown5612

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01 Aug 2025, 04:08

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KarishmaB, why did you subtract, 6c2, shouldn't it be added?
1- {(6C1*10C2 + 6C2) / 12C4}

Also, won't 6c1*10c2 already contain two couple possibilities as well? So why are we adding 6c2 to it? Getting confused here - not able to understand where I am going wrong in my thought process..

Also, why is my following method wrong:
1-{(6c2 + 6c1*10c1*8c1)/12c4}

6c2 being selecting 2 couples out of 6. 6c1*10c1*8c1 being selecting 1 couple and the other two being not couples (if one couple is selected, there are 10 people, so select 1 from there, then select one more person from remaining 4 couple groups, ie from 8 people)

Thanks in advance!

KarishmaB

kll08

Bunuel KarishmaB

Would it be possible to solve it using the following approach? However, I am arriving at a wrong answer even though logically it makes sense:

Probability (None) = 1 - Probability (All)

P(all) = 6C2 (Selection Two Couples) * 2C2 (Select both members of first couple) * 2C2 (Select both members of second couple) / 12C4
P(all) = 2 /33

P (none) = 1 - 2/33 = 31/33

Can you please help me understand on which step am I faltering?

Bunuel

Official Solution:

If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

A. \(\frac{1}{33}\)
B. \(\frac{2}{33}\)
C. \(\frac{1}{3}\)
D. \(\frac{16}{33}\)
E. \(\frac{11}{12}\)

Initially, there are 12 individuals in the group, and we need to choose one person to be the first member of the committee. Since any of the 12 individuals can be chosen, there is no restriction on our choice for the first member. Therefore, we can say that the probability of selecting any individual for the first member of the group is 1, as there is no constraint on this initial selection.

After selecting the first person, there are 11 people left to choose from for the second member, and since we want to avoid picking the spouse of the first person, we have 10 valid choices out of the 11 remaining individuals. The probability that the second person chosen is not married to the first person is thus \(\frac{10}{11}\).

For the third person, we need to ensure this individual is not married to either of the first two, so there are 8 valid choices out of the 10 remaining individuals. Therefore, the probability that the third individual is not married to either of the first two is \(\frac{8}{10}\).

Likewise, for the fourth person, we need to ensure that this individual is not married to any of the first three, so there are 6 valid choices out of the 9 remaining individuals. Consequently, the probability that the fourth individual is not married to any of the first three is \(\frac{6}{9}\).

Therefore, to find the probability of selecting four people with none of them married to each other, we multiply the probabilities of each step: \(P=1*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\).

Alternative approach:

Each couple can send only one "representative" to the committee. We can select 4 couples (since there should be 4 members) to send one representative each to the committee in \(C^4_6\) ways.

However, each of these 4 chosen couples can send either the husband or the wife, resulting in \(2*2*2*2=2^4\) possibilities.

So, the number of ways to choose 4 people from 6 married couples such that none of them are married to each other is: \(C^4_6*2^4\).

The total number of ways to choose 4 people out of 12 is \(C^4_{12}\).

The probability is \(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\)

Answer: D

We are not allowed the case where 2 couples are selected so you subtracted that out of 1.

How about the case where exactly 1 couple is selected along with 2 other people? We need to subtract that out too.

So we calculate the probability of selecting at least 1 couple as

\(\frac{(6C1*10C2 - 6C2) }{ 12C4} = \frac{17}{33} \)

KarishmaB

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02 Aug 2025, 03:23

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Assume couples: {H1, W1}. {H2, W2}, {H3, W3}.... etc.

6C1 gives us the number of ways in which we select a couple. Say I select {H2, W2}.
10C2 gives the number of ways of selecting 2 other people say {H1, W4} or {H3, W3} etc.

So 6C1 * 10C2 includes all cases in which exactly one couple is selected and in which two couple are selected. BUT the number of ways in which 2 couples are selected (6C2) is included twice here.
Why? Because 6C1 * 10C2 includes the way in which I selected {H2, W2} in 6C1 and {H3, W3} in 10C2 and it also includes the way in which I selected {H3, W3} in 6C1 and {H2, W2} in 10C2.

That is why we need to subtract out 6C2.

Rahuljaggu

KarishmaB