Question

You take a trip by air that involves three independent flights. If there is an 67% chance each specific leg of the trip is on time, what is the probability all three flights arrive on time

Answer 1

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

$P(X=3)=^3C_3(0.67)^3(0.33)^0\\\\P(X=3)=0.301$

Hence, the required probability is 0.301.