Three married couples have purchased concert tickets and are seated in a row con

Question

Three married couples have purchased concert tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random order…

use the multiplication rule to compute the probability that Adam and Jill sit together on the far left (event A) and that Bob and Mary (husband and wife) sit together in the middle (event B).

Given that Bob and Mary sit together in the middle, what is the probability that the two other husbands sit next to their wives?

Given that Bob and Mary sit together, what is the probability that all husbands sit next to their wives?

Explanation / Answer

When 6 persons i.e. 3 couples arranged in 6 seats total possible permutations possible are 6! = 720

a) If Adam and Jill sit together on the far left, remaining 4 persons can be arranged in 4! ways and Adam and Jill can be arranged in 2! ways

P(A) = 4!*2!/6! = 0.0667

b) If Bob and Mary sit together in the middle remaining 4 persons can be arranged in 4! ways and Bob and Mary can be arranged in 2! ways

P(B) = 4!*2!/6! = 0.0667

c) If Bob and Mary sit together in the middle remaining 2 couples can be sit on the either side, hence 2! and couples can be arranged among themseleves in 2!*2! ways. Also Bob and Mary can be arranged in 2! ways.

Required Probability = 2!*2!*2!*2!/6! = 0.0222

d) There are 3 couples, and can be arranged in 3! ways. And each couple can be among themseleves in 2! ways

Required probability = 3!*2!*2!*2!/6! = 0.0667

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