Question

According to the Department of Transportation, 27% of domestic flights were delayed in 2007 at JFK airport. Five flights are randomly selected at JFK.

What is the probability that all five flights are on time?a) 0.2073b) 0.7927c) 0.8487d) 0.1513

Answer 1

Answer: a) 0.2073

Explanation:

Binomial distribution formula :

`P(X=x)=^nC_xp^x q^(n-x)` ,

where n= total number of trials.

x= number of successes.

p= probability of getting success in each trial.

q= 1-p = probability of getting failure in each trial.

For the given question , let the flight are on time be success.

Let x denote the number of flight are on time .

Given : Proportion of domestic flights were delayed in 2007 at JFK airport : q=0.27

Proportion of domestic flights were on time in 2007 at JFK airport :

p=1-0.27=0.73

Five flights are randomly selected at JFK.

i.e. n=5

Now, the probability that all five flights are on time will be :-

`P(X=5)=^5C_5(0.73)^5 (0.27)^(5-5)\\\\=(1)(0.73)^5=0.2073071593\approx0.2073`

Hence, the probability that all five flights are on time = 0.2073