1. Air Canada estimates that 1% of its customers with purchased tickets fail to
ID: 3178605 • Letter: 1
Question
1. Air Canada estimates that 1% of its customers with purchased tickets fail to show up for their flights. For one particular flight, the plane has 300 seats and the flight has been fully booked. How many additional tickets can the airline sell so that there is at least a 90% chance that everyone who shows up will have a seat?
2.A life insurance insured 10000 individuals aged 30. The probability that a 35-year old will die within one year is 0.0035. Within the next year, what is the probability that the insurance company will pay between 30 and 33 claims (both inclusive) among these 10000 people?
3.In the last four months, I had to replace five light bulbs in my house. What are the chances that I won’t need to change a lightbulb next month?
4.Two taxi arrive on average at a certain street corner for every 20 minutes. Three people are waiting at the street corner for taxi (assuming they do not know each other and each one will have his own taxi). Each person will be late for work if they do not catch a taxi within the next 15 minutes. (a) What is the probability that all three people will make it to work on time? (b) What is the probability that exactly one person will late for work? (c) What is the probability that at least one person will late for work?
Explanation / Answer
Solution:-
1) We can have book 1 additional tickets the airline sell so that there is at least a 90% chance that everyone who shows up will have a seat.
Probability that customers with purchased tickets fail to show up for their flights = 0.01
Probability that customers with purchased tickets fail to show up for their flights = 1 - 0.01 = 0.99
We will check with 1 extra ticket and 2 extra tickets
For total booking of 301 tickets
x = 300, number of showing up costumers.
p = 0.99
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x < 300, n = 301) = 0.9514
For total booking of 302 tickets
x = 300, number of showing up costumers.
p = 0.99
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x < 300, n = 302) = 0.8053
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