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

Total people: 9

Married couple: 2 specific people among them

Total ways to choose 5 people from 9:

\[ \text{Total} = \binom{9}{5} = 126 \]

✅ Case 1: Both are selected

We fix the married couple (2 people), then choose 3 more from remaining 7:

\[ \binom{7}{3} = 35 \]

✅ Case 2: Both are NOT selected

We remove both from the pool, then choose 5 from remaining 7:

\[ \binom{7}{5} = \binom{7}{2} = 21 \]

✅ Favorable outcomes:

\[ \text{Favorable} = 35 + 21 = 56 \]

✅ Probability:

\[ \text{Required Probability} = \frac{56}{126} = \frac{28}{63} = \frac{4}{9} \]

✅ Final Answer: $\boxed{\dfrac{4}{9}}$