Probability problem solved in two ways. Can someone please conclude which method to use in which case

Question

I found an elementary problem on probability that left me a bit confused-

6 married couples are standing in a room. If 4 people are chosen at random, then the chance that exactly one married couple is among the 4 is..?

How I approached it

Out of 6 married couples, choose one in $^6C_1$ ways. Out of remaining 5, choose 2 in $^5C_2$ ways. Now we have 4 people in hand. They can be matched in 4 ways such that no two married people are chosen. So the total no. of favourable selections = ${^6C_1}×{^5C_2}×4$. The sample space is $^{12}C_4$. Probability = $\frac{{^6C_1}×{^5C_2}×4}{^{12}C_4}$. This gives the answer. But this was my second attempt. Here's what I did in first one-

I tried to use Probability multiplication theorem-

Choose one person. Its probability =$\frac{12}{12}$. Choose his/her spouse $\frac{1}{11}$. Choose the third person $\frac{10}{10}$. And not his/her spouse $\frac{8}{9}$. Multiply them all to get $\frac{12×1×10×8}{12×11×10×9}$ and I thought I had the answer. But I later realised that in the above method I've fixed the order of selection. And therefore, to include all orders, I must multiply the above by $^4C_2$ and I had my answer.

If you've read all of this, here's what I want to know - I don't know if any of the things I did above are correct. After doing this problem I am confused in almost every problem on which method to adopt. Can someone please conclude this..??

NOTE-I know the multiplication theorem can be used only for independent events. But it would be great if you give me an insight on that as well.

Thankyou

Answer 1

Well, I am giving two approaches using combinations and permutations which might clarify things

$\underline{Approach\; 1}$

• Select one pair: $\binom61$

• Select two of the remaining pairs, and choose one from each pair : $\binom52\cdot2^2$

• $Pr = \dfrac{\binom61\binom52\cdot2^2}{\binom{12}4} = \dfrac{48}{99}$

$\underline{Approach\;2}$

In the line up of $4$ chosen people, line up the chosen pair in $\binom61\cdot 4\cdot3 = 72$ ways, and for the remaining people,ensure in the numerator that no other pair is selected, thus

$\dfrac{72\cdot10\cdot8}{12\cdot11\cdot10\cdot9} = \dfrac{48}{99}$

The $1^{st}$ approach is mainly combination oriented, and the $2^{nd}$, mainly permutation oriented.

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Added a different example to clear your confusion

We are dealing with drawing w/o replacement, e.g. the probability of drawing $2$ red and $3$ blue balls from a pool of $5$ red and 4$ blue balls.

By the combination approach, we would use $\dfrac{\binom52\binom43}{\binom95}$

Using the multiplication rule, $P(RRBBB) = \frac59\frac48\frac47\frac36\frac25$, but we would need to multiply it by $\binom52$ because the two reds could occupy any two of the five positions. [ But this mulriplication factor is all too often forgotten by students]

I would advise that you use direct multiplication of probabilities when a specific order is given, and combinations otherwise

Answer 2

What you've done is quite correct. The probability of choosing, in order, any person, then their spouse, then any other person, then anyone except the third person's spouse is $1\times\frac1{11}\times1\times\frac89=\frac8{99}$, and this is precisely the probability of the first and second people being the only married couple in the four.

Now the probability of having exactly one married couple is the probability that the first and second people are the only married couple plus the probability that the first and third are the only married couple plus ... plus the probability that the third and fourth are the only married couple. This is $\binom 42=6$ probabilities. This is because if there is a single married couple then they are in some two positions, so exactly one of those things happens. All the probabilities are equal (your married couple is equally likely to be in any of the possible positions), so multiplying your original probability by $\binom 42$ gives the answer, as you did.