Question

A committee of five is to be chosen from a group of 8 people which include a married couple. The probability for the selected committee which may or may not have the married couple.

Answer 1

Hint: Here in the question, we need to determine the probability for the selected committee which may or may not have the married couple such that the committee should include a total of 5 members from the group of 8 members. For this, we will first evaluate the total number of ways of selecting married couples and not selecting a married couple and simultaneously the required probability.

Complete step by step answer:
We are provided with the data that total number of people in the group are 8
And, the committee consists of 5 people.
Case (i): When in the selected committee the married couple is included.
1. Now, total number of people from which we have to choose will be,
Since the couple is already included in the selected committee = 8-2 = 6
i.e. m (Total people) = 6
2. Since the married couple is already included in the committee, the total number of people who are supposed to be selected will be = 5-2 = 3
i.e. n = 3
Now, people who are to be selected for the committee will be
$
  ^6{c_3} = \dfrac{{6!}}{{3! \times 3!}} \\
   = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3!}} \\
   = \dfrac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \\
   = 5 \times 4 \\
   = 20 \\
 $
Total number of ways in which the committee can be selected when the couple is already included in the selected committee = 20
Case (ii): When in the selected committee, the married couple is not included
3. Since the married couple is not selected in the committee, the total number of people out of which we have to select the committee now will be = 8-2 = 6
i.e. m(Total people) = 6
4. Since the married couple will not be selected for the committee, the people who are supposed to be selected = 5
i.e. n = 5
5. Now, putting the values in the equation we will get
$
  ^6{c_5} = \dfrac{{6!}}{{5! \times (6 - 5)!}} \\
   = \dfrac{{6 \times 5!}}{{5! \times 1!}} \\
   = 6 \\
 $
Total number of ways in which the committee can be selected when the couple not included in the selected committee = 6
Hence, the total number of ways for the selected committee which may or may not have the married couple is given as $20 + 6 = 26$
Now, the total number of ways of selecting 5 members from a group of 8 members is given as
$
  ^8{c_5} = \dfrac{{8!}}{{5! \times 3!}} \\
   = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5! \times 3 \times 2 \times 1}} \\
   = 56 \\
 $
So, the probability for the selected committee which may or may not have the married couple is given as
$\dfrac{{26}}{{56}} = \dfrac{{13}}{{28}}$

Note: The probability of selecting ‘n’ number of people from a group of ‘m’ number of people is given by:
$^m{c_n} = \dfrac{{m!}}{{n!(m - n)!}}$
The ‘m’ in the question is selected to be 6 not 8 because, in both of the cases the couple is either selected or not selected which means in either of the cases from the main group 2 people are not selected.