AP Statistics Probability Question Traffic data revealed that 35 percent of automobiles traveling along a portion of an interstate highway were exceeding the legal speed limit. Using highway cameras and licence plate registrations, it was also determined that 52 percent of sport cars were also speeding along the same portion of the highway. What is the probability that a randomly selected car along the same portion of highway is a speeding sports car? 
The answer seems to be .182 rather than "It cannot be determined from the information given. They seem to have gotten .182 from multiplying .35*.52.
But that seems to be only the right answer if the question said that .52 of the speeding cars were sports cars. It may be that sports cars account for only .001 of the cars on the highway. What is the logic behind this calculation?
 A: 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.
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