Question

Of all airline flight requests received by a certain discount ticket broker, 60% are for domestic travel (D) and 40% are for international flights (I). Let x be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of x. (Hint: One possible outcome is DID, with the probability (.6)(.4)(.6) = .144)
Value of x Probability
0
1
2
3

Answer 1

Answer:

We have a binomial probability distribution, in which:

P(X = 0) = 0.064

P(X = 1) = 0.288

P(X = 2) = 0.432

P(X = 3) = 0.216

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it is for domestic travel, or it is for international travel. Flights are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Let x be the number of requests among the next three requests received that are for domestic flights.

3 flights, and 60% are for domestic travel. This means that $n = 3, p = 0.6$. The probabilities are given by:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.6)^{0}.(0.4)^{3} = 0.064$

$P(X = 1) = C_{3,1}.(0.6)^{1}.(0.4)^{2} = 0.288$

$P(X = 2) = C_{3,2}.(0.6)^{2}.(0.4)^{1} = 0.432$

$P(X = 3) = C_{3,3}.(0.6)^{3}.(0.4)^{0} = 0.216$