Question

April, Bill , Candace, and Bobby are to be seated at random in a row of 5 chairs. What is the probability that April and Bobby will occupy the seats at the end of the row?

Answer 1

There are 5 chairs and 4 persons.

Total number of ways(N) the can be sit is,

$N=^5p\ _4=\frac{5!}{1!}=120$

Ways(n) of April and Bobby will occupy the seats at the end of the row is,

$n=2*^3p\ _2=2*6=12$

Therefore,the probability that April and Bobby will occupy the seats at the end of the row,

$P=\frac{n}{N}=\frac{12}{120}=0.1$

Answer is 0.1