Question

Assume that the car lot contains 20 percent Lincolns, 45 percent Porches, and 35 percent BMWs. Of the Lincol?

Answer 1

Answer:

0.3261 = 32.61% probability that it is a Lincoln

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.

In this question:

Event A: Has two airbags and is white.

Event B: It is a Lincoln.

Probability of a car having two airbags and being white.

This is:

70%*70% of 20%(Lincolns).

60%*40% of 45%(Porches).

90%*30% of 35%(BMWs). So

$P(A) = 0.7*0.7*0.2 + 0.6*0.4*0.45 + 0.9*0.3*0.35 = 0.3005$

Probability of a car having two airbags and being white, and being a Lincoln:

70%*70% of 20%(Lincolns).

So

$P(A \cap B) = 0.7*0.7*0.2 = 0.098$

If the car has two airbags and is white, what is the probability that it is a Lincoln?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.098}{0.3005} = 0.3261$

0.3261 = 32.61% probability that it is a Lincoln