# Discrete Random Variables and Probability Distributions

1. Feb 8, 2010

### exitwound

1. The problem statement, all variables and given/known data

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function of Y appears in the accompanying table.

$$\begin{tabular}{ c | c c c c c c c c c c c} y & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\ \hline p(y) & .05 & .10 & .12 & .14 &.25 &.17 & .06 & .05 & .03 & .02 & .01 \\ \end{tabular}$$

a.) What is the probability that the flight will accommodate all ticketed passengers who show up?

3. The attempt at a solution

I don't know how to start this problem.

2. Feb 8, 2010

### korican04

Isn't it just the P(Y=<50)

3. Feb 8, 2010

### exitwound

I'm asking you. I don't know. I don't have an answer to this one so I don't know how to proceed.

4. Feb 8, 2010

### korican04

It asks what's the probability that all ticketed passengers who show up will get a seat. So if 51,52, 53,54 or 55 passengers show up then there are some who don't get seats.
So you have to compute what's the probability that 50 or less passengers show up.
If you add up the probabilities on this chart it doesn't add to 1.
So you can't just add the probabilities from Y=1 to Y=50.
So you have to compute the prob of Y=<50. which is also
1 - P(Y>50) = 1- (P(Y=51)+P(Y=52)+P(Y=53)+P(Y=54)+P(Y=55))

5. Feb 8, 2010

### exitwound

If you add up the probabilities on the chart you do get 1.

And I'm confused by the problem anyway. All ticketed passengers WON'T get seated if 55 show up, or 54. They still have tickets and are overbooked so don't get on.

6. Feb 8, 2010

### korican04

Right, so you can add the probabilities up to 50. You still get the same answer.

7. Feb 8, 2010

### exitwound

I think i'm interpreting the wording incorrectly. Am I to assume that the passengers up to #50 are the ones who "deserve" to be on the plane? I'm completely lost on the wording of the problem. It says that 55 people HAVE tickets. there's no way they can accommodate 55 tickets on the flight.

8. Feb 8, 2010

### korican04

Oh I see what you are confused about. The table of values represent the probability of how many people actually show up. So if only 49 people out of the 55 tickets sold show up everyone will be accommodated. If 51 people show up, then not everyone will be accommodated.
Just because 55 people bought tickets, doesn't mean that all those people will will show up. I believe that's what the problem is trying to explain.

9. Feb 8, 2010

### exitwound

So in this case, as long as 50 people show up, all tickets will be granted for the flight? I'm looking for P(Y≤50) which is P(46)+P(47)+...P(50)?

The question for B) is actually "What is the probability that not all ticketed passengers who show up can be accommodated?" which would then be P(Y>50) = P(51)+(P52)+P(53)+P(54)+P(55)?

10. Feb 13, 2010

### exitwound

c.) If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

I don't know how to do this one either.

11. Feb 16, 2010

### korican04

The prob that First person on standby would get on: P(Y=<49) you would need one person or more to not show up.
The third would be P(Y=<47)