A team of 8 couples, (husband and wife) attend a lucky draw in which 4 persons picked up for a prize. Then, the probability that there is at least one couple is

$\frac{15}{39}$

We have,

Probability of selecting at least one couple for the prize = 1- Probability of not selecting any couple for the prize

$= 1-\frac{\left({^{16}C}_1×{^{14}C}_1×{^{12}C}_1×{^{10}C}_14!\right)}{^{16}C_4}$

$= 1-\frac{16×14×12×10}{16×15×14×13}=\frac{15}{39}$