An airline claims that there is a 0.10 probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight. This outcome is independent from flight to flight. Sam is a frequent flier who always purchases coach-class tickets.
For the following questions, pick the correct value rounded to the nearest hundredth.

What is the probability that Sam will be upgraded exactly 3 times in his next 20 flights?

What is the probability that Sam will be upgraded at most 4 times in his next 20 flights?

What is the probability that Sam’s first upgrade will occur on the third flight?

Answer:

b

Step-by-step explanation:

a) The probability that Sam will be upgraded exactly 3 times in his next 20 flights is 19.01 %

b) The probability that Sam will be upgraded at most 4 times in his next 20 flights is 95.68 %

c) The probability that Sam’s first upgrade will occur on the third flight is 0.0729

What is a Binomial Distribution?

The binomial distribution is a type of probability distribution that predicts the likelihood of obtaining one of two outcomes given a set of inputs. It summarizes the number of tries where each trial has the equal chance of producing the same result.

The formula for Binomial Distribution is given by

P ( x ) = [ n! / ( n - x )! x! ] pˣqⁿ⁻ˣ

where

n = number of trials

x = number of successes

p = probability of getting a success in one trial

q = probability of getting a failure in one trial

q = 1 - p

Given data ,

Let the probability of getting a coach-class ticket be p

The value of p = 0.10

a)

The probability that Sam will be upgraded exactly 3 times in his next 20 flights

The number of trials = 20

The probability of success p = 0.10

The probability of failure q = 1 - p

The probability of failure q = 0.90

x = 3

Substituting the values in the equation , we get

P ( 3 ) = [ 20! / ( 17 )! 3! ] ( 0.10 )³ ( 0.90 )¹⁷

P ( 3 ) = 0.1901

P ( 3 ) = 19.01 %

b)

The probability that Sam will be upgraded at most 4 times in his next 20 flights

The number of trials = 20

The probability of success p = 0.10

The probability of failure q = 1 - p

The probability of failure q = 0.90

x ≤ 4

Substituting the values in the equation , we get

P ( x ≤ 4 ) = [ 20! / ( 16 )! 4! ] ( 0.10 )⁴ ( 0.90 )¹⁶

P ( x ≤ 4 ) = 0.9568

P ( x ≤ 4 ) = 95.68 %

c)

The probability that Sam’s first upgrade will occur on the third flight

A = ( 0.9 ) ( 0.9 ) ( 0.9 ) ( 0.1 )

A = 0.0729

A = 7.29 %

Hence , the solution is calculated by binomial distribution

To learn more about binomial distribution click :

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