00:01
The scenario for this question is a small regional aircraft carrier accepted 17 reservations for a flight that has 14 seats.
00:11
We were told that nine of the reservations went to regular customers who we know will arrive for the flight.
00:16
So that means there is an extra eight reservations that went to non -regular customers.
00:23
And we're told that the probability of each of these eight non -regular customers showing up is 51 percent, or 0 .51.
00:32
And we're asked to find the probability that an overbooking occurs.
00:38
So let's first define a random variable x as the number of the non -regular customers out of eight you show up for the flight.
00:44
So here each of these eight customers can be thought of as bernoulli trials with two outcomes of interest, either show up for the flight or not.
00:52
And it's given in the question that they arrive independently.
00:56
And so the number of successes in a fixed number of independent bernoulli trials is a binomial a random variable.
01:04
So here we can say x is a binomial based on eight trials and probability of success on each trial .5 .1.
01:11
Now in order to have an overbooking, we need more than 14 customers in total to show up.
01:18
We have the nine regular customers who are showing up.
01:21
So we need more than five of the non -regular customers showing up.
01:26
So we want to find the probability that x is greater than five.
01:30
This would result in an overbooking.
01:33
This can be expressed as one minus, the probability that x is at most 5.
01:42
And we can use software such as excel to solve this.
01:45
So we have 1 minus the probability that x is at most 5.
01:48
In excel, we started computation with an equal sign.
01:51
We first have 1 minus...