Answer: The required probability is 0.301.
Step-by-step explanation:
Since we have given that
Probability that each specific leg of the trip is on time = 67%
Number of independent flights = 3
We will use "Binomial distribution" to find the probability that all three flights arrive on time.
Here p = 0.67
q = 1-0.67 = 0.33
So, the probability that all three flights arrive on time is given by
[tex]P(X=3)=^3C_3(0.67)^3(0.33)^0\\\\P(X=3)=0.301[/tex]
Hence, the required probability is 0.301.