Question

A committee of 5 is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is
A. \[\dfrac{1}{2}\]
B. \[\dfrac{5}{9}\]
C. \[\dfrac{4}{9}\]
D. \[\dfrac{2}{3}\]

Answer 1

Hint: Here the thing that is to be noted is in one condition a married couple is to be served together and in other is not serving together. But the number of people to be chosen are only 5 out of 9. Once that includes the married couple and at second time no married couple will serve together.

Complete step-by-step solution:
First we will check for all possible combinations.
That is if we chose 5 people out of 9 and without any condition then the possible outcomes will be,
\[9{C_5} = \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}}\]
\[9{C_5} = \dfrac{{9!}}{{5! \times 4!}}\]
This can be solved as,
\[9{C_5} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5! \times 4 \times 3 \times 2 \times 1}}\]
On cancelling 5! and calculating the other numbers,
\[9{C_5} = 9 \times 2 \times 7\]
\[9{C_5} = 126\]
These are all possible outcomes without any condition.
Now we will deal with the special cases:
Case 1:
If there is one married couple M1W1 in the committee. Now we have to select 3 more people from the remaining 7 people. So the outcomes will be,
\[7{C_3} = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} = \dfrac{{7 \times 6 \times 5 \times 4!}}{{3! \times 4!}}\]
On cancelling 4! And taking the value of 3!
\[7{C_3} = \dfrac{{7 \times 6 \times 5}}{6}\]
\[7{C_3} = 35\]
This is the answer for the committee that will have married couples serve together.
Case 2:
Now if the committee will not include the married couple. So there will be 7 people remaining and we will select 5 people from the remaining 7.
\[7{C_5} = \dfrac{{7!}}{{5!\left( {7 - 5} \right)!}} = \dfrac{{7 \times 6 \times 5!}}{{5! \times 2!}}\]
On cancelling 5! We get,
\[7{C_5} = \dfrac{{42}}{2} = 21\]
This is the answer for the committee that will not have married couples that serve together.
Now the probability will be given by,
\[P\left( {event} \right) = \dfrac{{35 + 21}}{{126}}\]
\[P\left( {event} \right) = \dfrac{{56}}{{126}}\]
On dividing by 14,
\[P\left( {event} \right) = \dfrac{4}{9}\]
Thus this is the correct answer,
Thus option C is correct.

Note: Here note that generally when two contrast events are there we subtract the possible outcomes of the other from 1 but in this case that is not the way because the possible outcomes with the condition is not necessarily the contract of the given condition. Because not all persons are selected at the time.
If only the number of outcomes is asked then no need of finding the probability.