probability that car is red given one classifies red car 99%

Question

So a company has 100 cars where 1 is red and 99 are blue. One car meets with an accident and someones sees it as red car. This person sees a red as red 99% of the time and blue cars as red 2% of the time. What's the probability that the car was red given this guy saw a red car?

I got 0.009. I think it's wrong because after watching this video, I realized that I should also consider false positives. So from that logic, I got 0.33. Which one is correct?

Answer 1

The probabilities of the car being red or blue and the onlooker seeing it as red are $0.01×0.99$ and $0.99×0.02$ respectively. Then the answer is the first probability divided by the sum of the two probabilities, or $\frac13$. The second answer is correct.

Answer 2

Formally, one should define the events: $R=\{\hbox{the car is red} \}$, $B=\{\hbox{the car is blue} \}$, $S=\{\hbox{the car is seen red} \}$. Define also :$S_R=\{\hbox{the car is red and seen red} \}$ and $S_R=\{\hbox{the car is blue and seen red} \}$.

The question is to determine $P(R|S)$.

Using the definition of conditional probability: $P(R|S)=P(R \cap S)/P(S)$.

Here $P(S)=P(S,R)+P(S,B)=P(S|R)P(R)+P(S|B)P(B)$, and $P(R \cap S)=P(S_R)=P(S|R)P(R)$

From the exercice: $P(S|R)=0.99$, $P(S|B)=0.02$, $P(A)=0.01$, $P(B)=0.99$.

Finally $$ P(R|S)=\frac{P(S|R)P(R)}{P(S|R)P(R)+P(S|B)P(B)}=\frac{1}{1+\frac{0.02 \times 0.99}{0.01 \times0.99}}=\frac{1}{3}. $$

Answer 3

A good approach to this kind of problem is to write out all the possible scenarios that may have happened, and give them each "points" based on their probabilities. Then divide by the total points to get the probability of each scenario. The reason I use the word "points" is because we don't know the total in our denominator yet, so it's not a percentage or a fraction.

In our problem we have:

1. [The car is red] AND [the person saw it as red]: [1%] * [99%] = 0.0099 points
2. [The car is red] AND [the person saw it as blue]: we throw this out because we were told the person saw it as red. 0 points.
3. [The car is blue] AND [the person saw it as red]: [99%] * [2%] = 0.0198 points
4. [The car is blue] AND [the person saw it as blue]: we reject this as well. 0 points.

Our total points add up to 0.0297. Scenario 1 earned 1/3 of those points and scenario 3 earned 2/3.

Now the question is: "what is the probability the car was red?". Add up the points of all scenarios which have this property. The car was red in both scenario 1 and 2, so we're looking for (0.0099 + 0) / 0.0297 = 1/3.

I like this approach because it's obvious how to apply it to almost any problem. The number of possible scenarios can quickly get out of hand, however, and then you want to start understanding Bayes' Theorem and the formal methods that Didier Felbacq is discussing. Those are ultimately just shortcuts for doing this same kind of analysis.