A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 48% chance, independently of each other. A) Find the probability that overbooking occurs.
B) Find the probability that the flight has empty seats.

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. 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.

Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.

This means that

A) Find the probability that overbooking occurs.

12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.

32.04% probability that overbooking occurs.

B) Find the probability that the flight has empty seats.

Less than 4 of the booked passengers arrive.

To make it easier, i will use

From a), P(X > 4) = 0.3204

40.79% probability that the flight has empty seats.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47