Question

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 atleast one couple is

• A. $\frac{11}{39}$
• B. $\frac{12}{39}$
• C. $\frac{14}{39}$
• D. $\frac{5}{13}$

Answer 1

The correct option is D $\frac{5}{13}$

Total People Attending: 16
Ways To Choose 4 From 16: 16 x 15 x 14 x 13/4! = 1820
Ways To Choose 4 Without Any Couples: 16 x 14 x 12 x 10/4! = 1120
Ways To Choose 4 That Include One Or More Couple: 1820 - 1120 = 700
Probability Of Choosing At Least One Couple: 700/1820 = 35/91=5/13