Return to Answer

The calculation is wrong and your comment is correct

Your comment, "it may be that sports cars account for only .001 of the cars on the highway" is correct.

Let $A$ be the event that a car is speeding, and let $B$ be the event that a car is a sports car. We know $P(A)=.35$ and $P(A \mid B) = .52$.

We want to know the probability that a car is a car AND a sports car. By def. conditional probability:

$$ P(AB) = P(A \mid B) P(B)$$

But we do not know $P(B)$. We also know that $ P(A) = P(A \mid B)P(B) + P(A \mid \neg B)( 1 - P(B))$ but we still have two unknowns but only one equation. The probability of speeding given the car is not a sports car must be consistent with the below equation, but that doesn't nail down P(B).

$$ P(A \mid \neg B) = \frac{.35 - .52 P(B)}{1 - P(B)}$$

For example, we may have the probability of a sports car $P(B)$ is .5 and the probability of speeding for non-sports cars is .18. Or $P(B)$ is .001 and the probability of speeding for non-sports cars is .3498.

The calculation is wrong and your comment is correct

Your comment, "it may be that sports cars account for only .001 of the cars on the highway" is correct.

Let $A$ be the event that a car is speeding, and let $B$ be the event that a car is a sports car. We know $P(A)=.35$ and $P(A \mid B) = .52$.

We want to know the probability that a car is a car AND a sports car. By def. conditional probability:

$$ P(AB) = P(A \mid B) P(B)$$

But we do not know $P(B)$. We also know that $ P(A) = P(A \mid B)P(B) + P(A \mid \neg B)( 1 - P(B))$ but we still have two unknowns but only one equation. The probability of speeding given the car is not a sports car must be consistent with the below equation, but that doesn't nail down P(B).

$$ P(A \mid \neg B) = \frac{.35 - .52 P(B)}{1 - P(B)}$$

For example, we may have the probability of a sports car $P(B)$ is .5 and the probability of speeding for non-sports cars is .18. Or $P(B)$ is .001 and the probability of speeding for non-sports cars is .3498.

Homework solutions, lecture notes, published papers, textbooks etc... are not infallible

Academics generally do their best to eliminate typos and errors, but not all are caught.

Part of the learning process is developing the capacity and the confidence to correctly determine on your own whether an argument or statement is correct.

The calculation is wrong and your comment is correct

Let $A$ be the event that a car is speeding, and let $B$ be the event that a car is a sports car. We know $P(A)=.35$ and $P(A \mid B) = .52$.

We want to know the probability that a car is a car AND a sports car.

$$ P(AB) = P(A \mid B) P(B)$$

But we do not know $P(B)$. We also know that $ P(A) = P(A \mid B)P(B) + P(A \mid \neg B)( 1 - P(B))$ but we still have two unknowns but only one equation. The probability of speeding given the car is not a sports car must be consistent with the below equation, but that doesn't nail down P(B).

$$ P(A \mid \neg B) = \frac{.35 - .52 P(B)}{1 - P(B)}$$

The calculation is wrong and your comment is correct

Your comment, "it may be that sports cars account for only .001 of the cars on the highway" is correct.

Let $A$ be the event that a car is speeding, and let $B$ be the event that a car is a sports car. We know $P(A)=.35$ and $P(A \mid B) = .52$.

We want to know the probability that a car is a car AND a sports car. By def. conditional probability:

$$ P(AB) = P(A \mid B) P(B)$$

But we do not know $P(B)$. We also know that $ P(A) = P(A \mid B)P(B) + P(A \mid \neg B)( 1 - P(B))$ but we still have two unknowns but only one equation. The probability of speeding given the car is not a sports car must be consistent with the below equation, but that doesn't nail down P(B).

$$ P(A \mid \neg B) = \frac{.35 - .52 P(B)}{1 - P(B)}$$

For example, we may have the probability of a sports car $P(B)$ is .5 and the probability of speeding for non-sports cars is .18. Or $P(B)$ is .001 and the probability of speeding for non-sports cars is .3498.

There isn't enough information to answer the question.

The calculation is wrong and your comment is correct

Let $A$ be the event that a car is speeding, and let $B$ be the event that a car is a sports car. We know $P(A)=.35$ and $P(A \mid B) = .52$.

We want to know the probability that a car is a car AND a sports car.

$$ P(AB) = P(A \mid B) P(B)$$

But we do not know $P(B)$. We also know that $ P(A) = P(A \mid B)P(B) + P(A \mid \neg B)( 1 - P(B))$ but we still have two unknowns but only one equation. The probability of speeding given the car is not a sports car must be consistent with the below equation, but that doesn't nail down P(B).

$$ P(A \mid \neg B) = \frac{.35 - .52 P(B)}{1 - P(B)}$$