Gmat_mission wrote: ↑

Sun Dec 27, 2020 3:52 pm

A dance group consists of \(6\) men and \(7\) women. There are \(3\) married couples in the group. If \(2\) people have to be selected \(-1\) man and \(1\) woman, what is the probability that one of the couples will be selected?

A. \(\dfrac1{14}\)

B.\(\dfrac17\)

C. \(\dfrac16\)

D. \(\dfrac13\)

E. \(\dfrac12\)

Answer: A

Solution:

The number of ways to choose 1 man and 1 woman from 6 men and 7 women, respectively, is 6C1 x 7C1 = 42. The number of ways to choose 1 married couple from 3 married couples is 3C1 = 3. Thus the probability is 3/42 = 1/14.

Alternate Solution:

The probability that the first married man and his wife are chosen is 1/6 x 1/7 = 1/42.
The probability that the second married man and his wife are chosen is also 1/6 x 1/7 = 1/42. And the probability that the third married man and his wife are chosen is also 1/6 x 1/7 = 1/42.

Thus, the probability that any one of these three married couples is chosen is 1/42 + 1/42 + 1/42 = 3/42 = 1/14.

Answer: A