A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 52% chance, independently of each other. (Report answers accurate to 4 decimal places.) Required:
a. Find the probability that overbooking occurs.
b. Find the probability that the flight has empty seats.

a) 0.2135 = 21.35% probability that overbooking occurs.

b) 0.4625 = 46.25% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passengers, there are only two possible outcomes. Either they arrive, or they do not. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

We are interest in the 19-14 = 5 remaining reservations, each of whom arrive with 52% probability

This means that

a. Find the probability that overbooking occurs.

There are 17-14 = 3 seats remaining. So overbooking occurs if more than 3 arrive.

21.35% probability that overbooking occurs.

b. Find the probability that the flight has empty seats.

3 seats remaining, so this is the probability that less than 3 passengers arrive.

0.4625 = 46.25% probability that the flight has empty seats.

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.