According to an airline, flights on a certain route are on time 8080% of the tim
ID: 3061169 • Letter: A
Question
According to an airline, flights on a certain route are on time 8080% of the time. Suppose 1010 flights are randomly selected and the number of on-time flights is recorded.
(a) Explain why this is a binomial experiment.
(b) Find and interpret the probability that exactly 6 flights are on time.
(c) Find and interpret the probability that fewer than 6 flights are on time.
(d) Find and interpret the probability that at least 6 flights are on time.
(e) Find and interpret the probability that between 4 and 6 flights, inclusive, are on time.
Explanation / Answer
a.)
Experiment is performed fixed number of time (n) = 10, each time called a trial.
The trials are independent.
Each trial has two possible outcomes, usually called a success (here success is flights are on time = 80% = 0.8) and a failure (here failure is flights not on time = 1 - probability of success = 1 - 0.8 = 0.2).
The probability of success is the same for every trial.
Since, these are the characteristics for binomial experiment, thus this is a binomial experiment.
b.)
According to Bernoulli Trials
Probability of getting r times success in n number of trials is equal to
= n C r * (Probability of success) r * (Probability of failure) (n - r)
Probability that exactly 6 flights are on time
= Probability of 6 success
= 10 C 6 * (0.8) 6 * (0.2) (10 - 6)
= 10 C 6 * (0.8) 6 * (0.2) 4
c.)
According to Bernoulli Trials
Probability of getting r times success in n number of trials is equal to
= n C r * (Probability of success) r * (Probability of failure) (n - r)
Probability that fewer than 6 flights are on time
= Probability of 0 success + Probability of 1 success + Probability of 2 success + Probability of 3 success + Probability of 4 success + Probability of 5 success
= 10 C 0 * (0.8) 0 * (0.2) (10 - 0) + 10 C 1 * (0.8) 1 * (0.2) (10 - 1) + 10 C 2 * (0.8) 2 * (0.2) (10 - 2) + 10 C 3 * (0.8) 3 * (0.2) (10 - 3) + 10 C 4 * (0.8) 4 * (0.2) (10 - 4) + 10 C 5 * (0.8) 5 * (0.2) (10 - 5)
= 10 C 0 * (0.2) 10 + 10 C 1 * (0.8) 1 * (0.2) 9 + 10 C 2 * (0.8) 2 * (0.2) 8 + 10 C 3 * (0.8) 3 * (0.2) 7 + 10 C 4 * (0.8) 4 * (0.2) 6 + 10 C 5 * (0.8) 5 * (0.2) 5
d.)
According to Bernoulli Trials
Probability of getting r times success in n number of trials is equal to
= n C r * (Probability of success) r * (Probability of failure) (n - r)
Probability that atleast 6 flights are on time
= Probability of 6 success + Probability of 7 success + Probability of 8 success + Probability of 9 success + Probability of 10 success
= 10 C 6 * (0.8) 6 * (0.2) (10 - 6) + 10 C 7 * (0.8) 7 * (0.2) (10 - 7) + 10 C 8 * (0.8) 8 * (0.2) (10 - 8) + 10 C 9 * (0.8) 9 * (0.2) (10 - 9) + 10 C 10 * (0.8) 10 * (0.2) (10 - 10)
= 10 C 6 * (0.8) 6 * (0.2) 4 + 10 C 7 * (0.8) 7 * (0.2) 3 + 10 C 8 * (0.8) 8 * (0.2) 2 + 10 C 9 * (0.8) 9 * (0.2) 1 + 10 C 10 * (0.8) 10
e.)
According to Bernoulli Trials
Probability of getting r times success in n number of trials is equal to
= n C r * (Probability of success) r * (Probability of failure) (n - r)
Probability that between 4 and 6 flights are on time
= Probability of 4 success + Probability of 5 success + Probability of 6 success
= 10 C 4 * (0.8) 4 * (0.2) (10 - 4) + 10 C 5 * (0.8) 5 * (0.2) (10 - 5) + 10 C 6 * (0.8) 6 * (0.2) (10 - 6)
= 10 C 4 * (0.8) 4 * (0.2) 6 + 10 C 5 * (0.8) 5 * (0.2) 5 + 10 C 6 * (0.8) 6 * (0.2) 4
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