Question

according to an airline, a particular flight is on time 87% of the time. suppose 39 flights are randomly selected and the number of on time flights is recorded. find the probabilities of the following events occurring.
a. all 39 flights are on time
b. between 33 and 35 flights (inclusive) are on time
a. the probability that all 39 flights are on time is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

For part (a), \(n = 39\), \(k=39\), \(p = 0.87\), and \(1-p=0.13\)

Step2: Calculate \(C(39,39)\)

\(C(39,39)=\frac{39!}{39!(39 - 39)!}=1\)

Step3: Calculate the probability

\(P(X = 39)=C(39,39)\times(0.87)^{39}\times(0.13)^{0}\)

Since \((0.13)^{0}=1\) and \(C(39,39) = 1\), we have \(P(X = 39)=(0.87)^{39}\)

Using a calculator, \((0.87)^{39}\approx0.0041\)

Answer:

\(0.0041\)