Question

Four couples are at a party. Four of the eight people are randomly selected to win a prize. No person can win more than one prize. What is the probability that both of the members of at least one couple win prizes? Express your answer as common fraction.

Answer 1

Answer:

27/35

Step-by-step explanation:

We use combination to solve for this

C(n, r), =nCr = n!/r!(n - r)!

From the question, we are told that:

Four couples are at a party. Four of the eight people are randomly selected to win a prize.

Four couples = 8 people.

= 8C4 = 8!/4! (8 - 4)!

= 70

No person can win more than one prize. ( No person can win more than one prize of the 4 people selected)

This can happen in 4 ways

[4C1 × 3C2 ] × 4=

[4!/1! ×( 4 - 1)!] × [3!/2! ×(3-2)!] × 4 ways

= 4 × 3 × 4 ways

= 48

The probability that both of the members of at least one couple win prizes

48 + 4C2/ 8C4

4C2 = 4!/2!(4 - 2) !

= 6

8C4 = 8C4 = 8!/4! (8 - 4)!

= 70

48 + 6/ 70

= 54/70

= 27/35

Therefore, the probability that both of the members of at least one couple win prizes is 27/35.