Question

Three married couples have purchased theater tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random fashion (random order), what is the probability that Jim and Paula (husband and wife) sit in the two seats on the far left?

Answer 1

The required probability is :
`(1)/(15)`

Explanation:

Three married couples have purchased theater tickets and are seated in a row consisting of just six seats.

First we will check the total arrangements that is 6! ways.

6! =
`6*5*4*3*2*1=720`

Jim and Paula can sit at far left in 2 ways and the remaining 4 in 4! ways,.

So, probability will be =
`2*(4!)/(6!)`

=
`2*(24)/(720)`

=
`(48)/(720)`

=
`(1)/(15)`