A ballroom contains four married couples (four men and their four wives). The ge

Question

A ballroom contains four married couples (four men and their four wives). The gentlemen are to be paired up randomly with the ladies, one man to each woman. After one dance, the eight people are to be separated, except for the married couples that happen to be together by chance. The separated people will then be paired up again randomly for the second dance, again one man to each woman.

(a) Find the probability that exactly two married couples will be dancing together on the second dance.

(b) Suppose that exactly one married couple are dancing together on the second dance. What is the probability that no man was dancing with his wife on the first dance?

Explanation / Answer

(a) Let P(M, M) = the probability that exactly M married couples were paired on the first pairing, and exactly M couples (excluding the M couples) were paired on the second pairing.

Then, the probability that exactly two married couples are dancing together on the second dance is:

P(2, 0) + P(1, 1) + P(0, 2)

Of the four men, the first may be paired with any of the 4 wives, the second with any of the 3 remaining wives, and so on, making 4! ways in which couples may be paired on the first pairing.

There are C(4, 2) = 6 ways in which exactly two married couples may be paired on the first pairing. The remaining two couples may be paired in two possible ways, one of which results in zero married couples being paired. . So:

P(2, 0) = C(4, 2) / 4! × C(2, 0) / 2 = 6 / 48 = 1 / 8


Choose any of the 4 married couples to be paired on the first pairing; each of these may occur along with the !3 = 2 derangements of the remaining three couples. On the second pairing, there are 3 ways for exactly one of the three remaining married couples to be paired, out of a possible 3! pairings. So:

P(1, 1) = (4 × !3 / 4!) × (3 / 3!) = 1 / 6


There are !4 = 9 ways in which none of the four married couples are paired on the first pairing. On the second pairing, there are C(4, 2) = 6 pairs resulting in exactly two married couples being paired, out of a possible 4! pairings. So,

P(0, 2) = (9 / 4!) × (6 / 4!) = 3 / 32


So,

P(2, 0) + P(1, 1) + P(0, 2)
= 1 / 8 + 1 / 6 + 3 / 32
= 37 / 96
0.39 SOLUTION PART A



(b) If exactly one married couple are paired on the second dance, then the (M, M) pairing was either (1, 0), or (0, 1). So the desired probability is:

P(0, 1) / [ P(0, 1) + P(1, 0) ]


Now, P(0, 1) = (!4 / 4!) × [ (4 × !3) / 4! ] = 1 / 8

And, P(1, 0) = [ (4 × !3) / 4! ] × (!3 / 3!) = 1 / 9


So, the desired probability is:

(1 / 8) / (1 / 8 + 1 / 9)

= 9 / 17

0.53 SOLUTION PART B

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