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

A
$11 / 39$
B
$12 / 39$
C
$14 / 39$
D
$15 / 39$

Solution:

We have,

Probability of selecting at least one couple for the prize

= 1 - Probability ofnot selecting any couple for the prize

$\begin{array}{l} = 1 - \frac{^{8} C_{4} {(2)}^{4}}{^{16} C_{4}} \\ = 1 - \frac{8 \times 7 \times 6 \times 5 \times 16}{16 \times 15 \times 14 \times 13} = \frac{15}{39} . \end{array}$

Solution:

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