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

Answer:

The required probability is : $\frac{1}{15}$

Step-by-step 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! = $6\times5\times4\times3\times2\times1=720$

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

So, probability will be = $2\times\frac{4!}{6!}$

= $2\times\frac{24}{720}$

= $\frac{48}{720}$

= $\frac{1}{15}$