Probability

Question

10 men and their wives participate in a corporate mixed-doubles tennis championship. What is the probability that no couple play in the second game?

• A. 0.6222
• B. 0.3111
• C. 0.4285
• D. 0.2777

Answer 1

A mixed doubles game has 4 players- 2 men and 2 women.

Number of ways to select men = ¹⁰C₂

Number of ways to select women = ¹⁰C₂

So, total number of ways = ¹⁰C₂ * ¹⁰C₂

Cases were no couples are there = ¹⁰C₂ * ⁸C₂ (select 2 men from all 10 possibilities and select 2 women from all possibilities excluding the 2 wives)

So, required probability = ¹⁰C₂ * ⁸C₂ / (¹⁰C₂ * ¹⁰C₂)
= 8*7/(10*9)
= 56/90
= 0.6222

Answer 2

we will have to choose 1 man and 1 woman for each side and also see that they are not husband-wife.  So, ¹⁰C₁ for choosing man and ⁹C₁ for woman. Similarly for other side we have to see that none of the players in the game are husband-wife . So, ⁸C₁ for man and ⁷C₁ for woman. We will have to permute all the players in a game. So, ¹⁰C₁ * ⁹C₁ * ⁸C₁ * ⁷C₁ * 4! . 
Total no. of matches to be played can be ¹⁰C₁ for man on one side and ⁹C₁ for man on other side. ¹⁰C₁ for woman on one side and ⁹C₁ for woman on other side. So, ¹⁰C₁ * ¹⁰C₁ * ⁹C₁ * ⁹C₁ * 4! . 
Probability= (¹⁰C₁ * ⁹C₁ * ⁸C₁ * ⁷C₁ * 4! ) / (¹⁰C₁ * ¹⁰C₁ * ⁹C₁ * ⁹C₁ * 4!)
                         = 0.6222

Comments

Sir...Why did you multiplied it by 4!?.....shouldn't it be 2! ...why do we need to permute 2 players of same team.....?

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Sorry i am wrong. It should be 3.
If we have 4 players (1,2,3,4), then matches can be between :
(1-2 vs 3-4)
(1-3 vs 2-4)
(1-4 vs 2-3)