SOLUTION: Because not all air line passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers.the probability that a passenger doesn

Question 1083199: Because not all air line passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers.the probability that a passenger doesn't show up is 0.10, and the passengers behave independently.
a/ what is the probability that every passenger who show up can take the flight?
b/ what is the probability that the flight departs with empty seats?
Found 2 solutions by rothauserc, Theo:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The Probability(P) that a passenger shows up is 1 - 0.10 = 0.90
:
We use the binomial probability formula
:
P(k successes out of n trials) = nCk * p^k * (1-p)^(n-k), where n is 125 and nCk = n! / (k! * (n-k)!)
:
Note that since the airline sold 125 tickets, then 125 passengers could show up,
so n=125
:
a) P ( X < or = 120) = 1 - P( k = 121) - P( k = 122) - P(k = 123) - P(k = 124) - P(k = 125) = 1 - 0.004 = 0.996
:
b) P ( X < 120 ) = 1 - P( X > or = 120) = 1 - 0.011 = 0.989
:

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
125 bought tickets.
only 120 seats available.

the probability that a person who bought a ticket will not show up is .1.

therefore, the probability that a person who bought a ticket does show up is 1 minus .1 which is equal to .9.

if less than 121 people show up, then everybody gets a seat.
that's because 120 or less show up and there are 120 seats available.

if less than 120 people show up, then there will be at least one empty seat.
that because 119 or less show up and there are 120 seats available.

when doing the calculations manually, you want to exert as little effort as possible.

the probability that less than 121 people show up is equal to 1 minus the probability that 121 or more show up.

the probability that 121 or more people show up is equal to the sum of p(121) + p(122) + p(123) + p(124) + p(125).

the binomial probability formula is p(x) = c(n,x) * p^x * q^(n-x)

x is the number of successes and can range from 0 to 125 in this problem.
p is the probability of success which is .9 since the probability that a person will show up is .9.
q is the probability of failure which is .1 since the probability that a person will not show up is .1.
c(n,x) is the number of ways you can get sets of x elements each out of a set of n elements.
that formula for c(n,x)is c(n,x) = n! / (x! * (n-x)!)

i used excel to do the calculations.

an example of one of the calculations would be p(121) = c(125,121) * .9^121 * .1^4.

c(125,121) is equal to 9691375
.9^121 is equal to 2.906321 * 10^-6
.1^4 is equal to 1 * 10^-4

p(121) is therefore equal to 9691375 * 2.906321 * 10^-6 * 1 * 10^-4.
the result is p(121) = .002816625.

a picture of the excel printout, shown below, confirms this.

the probability that everybody who shows up will get a seat is the same as the probability that less than 121 people will show up.

the probability that less than 121 people show up is equal to 1 minus the the probability that 121 or more people show up.

the probability that 121 or more people show up is the sum of p(121) through p(125) as shown in the following excel printout.

the sum of p(121) through p(125) is equal to 0.0039 rounded to 4 decimal places as shown in that same printout.

1 minus the sum of p(121) through p(125) is equal to 0.9961 rounded to 4 decimal places as also shown in that same printout.

the probability that there will be at least one empty seat is the probability that less than 120 people will show up.

this is equal to 1 minus the probability that 120 or more people will show up.

the probability that 120 or more people show up is the sum of p(120) through p(125) as shown in the following excel printout.

the sum of p(120) through p(125) is equal to 0.0114 rounded to 4 decimal places as shown in that same printout.

1 minus the sum of p(120) though p(125) is equal to 0.9886 rounded to 4 decimal places as also shown in that same printout.

you were asked the following questions.

a/ what is the probability that every passenger who show up can take the flight?

the answer to this is that the probability that every passenger who shows up can take the flight is .9961 or 99.61%.

b/ what is the probability that the flight departs with empty seats?

the answer to this is that the probability that the flight departs with empty seats is .9886 or 98.86%.

this all assumes that i did this correctly, which i believe i did, but there's always the probability that i thought i did it correctly but actually didn't.

i'm guessing i did it correctly this time.

you can also use an online binomial calculator to provide these results, and to confirm that you did it correctly.

one such calculator can be found here.

http://onlinestatbook.com/2/calculators/binomial_dist.html

the main this is to make sure that your assumptions are correct.

using this calculator, i did the following:

n is equal to 125 because that's the number of tickets that were sold.

p is .9 because that's the probability that a person who bought a ticket will show up for the flight.

if the number of people who show up is less than 121, then all of the people who show up will get a seat because 120 or less will show up and there are 120 seats on the plane.

put 121 in the less than field and click recalculate and the calculator will tell you that the probability is .9961.

if the number of people who show up is less than 120, then there will be at least 1 empty seat.

that's because 119 or less show up and there are 120 seats available.

put 120 in the less than field and click recalculate and the calculator will tell you that the probability is .9886.

this agrees with what i calculated in excel and took a lot less work in setting up that i had to when i did the calculations in excel.

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