Answer:
The probability that all the five flights are delayed is 0.2073.
Step-by-step explanation:
Let X = number of domestic flights delayed at JFK airport.
The probability of a domestic flight being delayed at the JFK airport is, P (X) = p = 0.27.
A random sample of n = 5 flights are selected at JFK airport.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={5\choose x}0.27^{x}(1-0.27)^{5-x};\ x=0,1,2...[/tex]
Compute the probability that all the five flights are delayed as follows:
[tex]P(X=5)={5\choose 5}0.27^{5}(1-0.27)^{5-5}=1\times 1\times 0.207307=0.2073[/tex]
Thus, the probability that all the five flights are delayed is 0.2073.