1 Answer

Clementine · Answer 1 · 2020-08-11T04:15:47+0000

Answer:

There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.

Explanation:

Binomial probability distribution

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

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

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

C_(n,x) = (n!)/(x!(n-x)!)

And
$\pi$ is the probability of X happening.

For this problem,

There are 560 tickets, so
n = 560

There is a 1% probability that a passenger will not show up for the flight, so
$\pi = 0.01$

How likely is it that there are more passengers showing up for the flight than seats are available?

An airline has sold 560 tickets for a certain flight (with capacity 555 seats).

So, this question is: What is the probability that at most 4 passengers do not show up for the flight.

P = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

P(X = 0) = C_(560,0).(0.01)^(0).(0.99)^(560) = 0.0036

P(X = 1) = C_(560,1).(0.01)^(1).(0.99)^(559) = 0.0203

P(X = 2) = C_(560,2).(0.01)^(2).(0.99)^(558) = 0.0574

P(X = 3) = C_(560,3).(0.01)^(3).(0.99)^(557) = 0.1079

P(X = 3) = C_(560,4).(0.01)^(4).(0.99)^(556) = 0.1517

So

P = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0036 + 0.0203 + 0.0574 + 0.1079 + 0.1517 = 0.3409

There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.

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