Question

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:
A. $\dfrac{{16}}{{33}}$
B. $\dfrac{8}{{33}}$
C. $\dfrac{{17}}{{33}}$
D. $\dfrac{{24}}{{33}}$

Answer 1

This is a question of combination. Since we are not talking about order here, it is a question of combination. If we talk about order then it will be a question of permutation.

Formula used: $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$probability = \dfrac{{possible{\text{ }}outcome}}{{{\text{total }}outcome}}$

Complete step by step solution: In this question six married couples are given,
Hence total persons are $6 \times 2 = 12$ persons
Out of these 12 persons, four people are chosen at random, so total outcomes are:
$^{12}{C_4} = \dfrac{{12!}}{{4!\left( {12 - 4} \right)!}}$
$^{12}{C_4} = \dfrac{{12!}}{{4!8!}}$
$^{ \Rightarrow 12}{C_4} = \dfrac{{12 \times 11 \times 10 \times 9 \times 8!}}{{4 \times 3 \times 2 \times 1 \times 8!}}$
$^{12}{C_4} = 495$
[ $8!$ Is cancel out from numerator and denominator]
Now we will calculate our favourable case:
We have to choose 4 people out of which one must be a married couple.
So possible ways for one married couple out of 6 couples are:
$^6{C_1} = \dfrac{{6!}}{{1!(6 - 1)!}}$
$^6{C_1} = \dfrac{{6!}}{{1!5!}}$
${ \Rightarrow ^6}{C_1} = \dfrac{{6 \times 5!}}{{1 \times 5!}}$
$^6{C_1} = 6{\text{ }}ways$
Now we will select the remaining 2 people out of the remaining 10 people who are not a married couple.
Our possibilities are:
Select 2 male from remaining five male i.e.
$^5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}$
${ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{2!3!}}$
$^5{C_2} = \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}}$
$^5{C_2} = 10{\text{ }}ways$
Select 2 females from remaining five females i.e.
$^5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}$
${ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{2!3!}}$
$^5{C_2} = \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}}$
$^5{C_2} = 10{\text{ }}ways$
Select 1 male and 1 female from remaining people i.e. [ Here we use 4 in one part because if I write 5 instead of 4 then the possibility will be a married couple again, so if I take 5 male then I have to skip wife of one male]
$^5{C_1}{ \times ^4}{C_1} = 5 \times 4 = 20{\text{ }}ways$
Hence Favourable outcomes are:
$6 \times (10 + 10 + 20) = 240{\text{ }}ways$
Now our outcome is given as,$outcome = \dfrac{{Possible{\text{ }}outcome}}{{{\text{total }}outcome}}$
i.e. $outcome = \dfrac{{240}}{{495}} = \dfrac{{16}}{{33}}$

Option (A) is our correct option.

Note: Students take care of people to be chosen. Read the statement carefully and apply the condition properly. Many students get confused with how to choose a favourable case. Students generally make mistakes in selecting the correct case for given situations. Make sure to check if ordering matters in these types of questions because that will be responsible for applying the formulas.