Answer:
A. 7.88*10^(-16)
B. 1.0000000 (almost certainty)
Explanation:
United Airlines flight 1832 from chicago to orlando is on time 80% of the time, according to united states airlines. Suppose 65 flights are randomly selected.
Binomial distribution will be required for the discrete case where there can be only two outcomes for each trial, success or failure, and where the number of experiments is known, and the probability of success is known and remains constant.
P(x) = n* (C(n,x)*p^x*(1-p)^(n-x)
where
n = size of experiment in number of trials
p = probability of success of one individual trial
x = number of successes
C(n,x) = number of combinations when x objects are taken out of n
= n!/(x!*(n-x)!)
P(x) = probability of success out of n experiments.
A) find the probability that exactly 22 flights are on time.
n = 65
x = 22
p = 0.80
P(22) = C(65,22) (0.80^22 * 0.20^(65-22))
= 65!/(22!(65-22)!) * (0.80^22 * 0.20^(65-22))
= 7.88*10^(-16)
B) Find the probability that at least 2 flights are on time.
We find the probabilities of 0 and 1 flight on time, and subtract from 1 to get probability of at least 2 on time.
P(0) = C(65,0)*0.8^0*0.2^65 = 3.689*10^(-46)
P(1) = C(65,1)*0.8^1*0.2^64 = 9.59*10^(-44)
Therefore probability of having at least 2 flights on time equals
1-P(0)-P(1) = 1-3.689*10^(-46)-9.59*10^(-44) = 1.0 (almost certainty)