Mathematics, 29.07.2021 01:00, Savageman4654

Of the travelers arriving at a small airport, 60% fly on major airlines, 20% fly on privately owned planes, and the remainder fly on commercially owned planes not belonging to a major airline. Of those traveling on major airlines, 50% are traveling for business reasons, whereas 70% of those arriving on private planes and 80% of those arriving on other commercially owned planes are traveling for business reasons. Suppose that we randomly select one person arriving at this airport. What is the probability that the person
a. is traveling on business?
b. is traveling for business on a privately owned plane?
c. arrived on a privately owned plane, given that the person is traveling for business reasons?
d. is traveling on business, given that the person is flying on a commercially owned plane?

Answers: 1

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Answers

answered: rexard

a) 0.55 = 55% probability that the person is traveling on business

b) 0.14 = 14% probability that the person is traveling for business on a privately owned plane.

c) 0.2545 = 25.45% probability that the person arrived on a privately owned plane, given that the person is traveling for business reasons.

d) 0.2 = 20% probability that the person is traveling on business, given that the person is flying on a commercially owned plane.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Question a:

50% of 60%(major airlines)

70% of 20%(privately owned airplanes)

80% of 100 - (60+20) = 20%(comercially owned airplanes). So

p = 0.5*0.5 + 0.7*0.2 + 0.8*0.2 = 0.55

0.55 = 55% probability that the person is traveling on business.

Question b:

70% of 20%, so:

p = 0.7*0.2 = 0.14

0.14 = 14% probability that the person is traveling for business on a privately owned plane.

Question c:

Event A: Traveling for business reasons.

Event B: Privately owned plane.

0.55 = 55% probability that the person is traveling on business.

This means that P(A) = 0.55

0.14 = 14% probability that the person is traveling for business on a privately owned plane.

This means that $P(A \cap B) = 0.14$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.14}{0.55} = 0.2545$

0.2545 = 25.45% probability that the person arrived on a privately owned plane, given that the person is traveling for business reasons.

Question d:

Event A: Commercially owned plane.

Event B: Business

80% of those arriving on other commercially owned planes are traveling for business reasons.

This means that:

P(B|A) = 0.2

0.2 = 20% probability that the person is traveling on business, given that the person is flying on a commercially owned plane.

answered: Guest

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05

answered: Guest

i think it would be 42cm