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. \( \frac{5}{9} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{9} \)

Hint:

When solving probability problems involving combinations, break the problem into cases and calculate the favorable outcomes for each case.

Answer 1

The Correct Option is D

We are asked to find the probability that a certain married couple will either serve together or not at all on a committee of 5 chosen from a group of 9 people. 
Step 1: Total number of ways to select the committee. 
We need to select 5 people from a group of 9. The total number of ways to do this is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of people and \( r \) is the number of people to be chosen: \[ \binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126. \] Step 2: Case 1 - The couple serves together. 
If the couple serves together, we treat them as a single unit, so we are left with selecting 3 additional members from the remaining 7 people. The number of ways to do this is: \[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35. \] Step 3: Case 2 - The couple does not serve. 
If the couple does not serve, we need to select all 5 members from the remaining 7 people. The number of ways to do this is: \[ \binom{7}{5} = \frac{7 \times 6}{2 \times 1} = 21. \] Step 4: Total favorable outcomes. 
The total number of favorable outcomes is the sum of the favorable outcomes from both cases: \[ 35 + 21 = 56. \] Step 5: Probability. 
The probability is the ratio of favorable outcomes to total outcomes: \[ \frac{56}{126} = \frac{4}{9}. \] Thus, the probability that the married couple will either serve together or not at all is \( \boxed{\frac{4}{9}} \).