# Probabiility problem. Maybe a conditional problem?

A car committed hit-and-run. It is fact that the car is either black or white.

Apparently, among all the black and white cars in the city, 70% are black and 30% are white.

A witness testified that the car was white. However, because he witnessed in a distance, there is only 70% that he can identify colors correctly.

What is the Probability that the car is in fact, white?

At first I thought the answer was P(White|witness correct)=$$\frac{\frac {30}{100}\times\frac{70}{100}}{\frac{70}{100}}$$ which would give probability of 30%. But I'm not sure if that's the right way of doing it.

## 2 Answers

This is a classic Bayes' Theorem problem. I think the simplest way to visualize it is to draw out a tree diagram with two levels: the top level is that the car is actually black or actually white. For each of these options, there are two options: the witness says they saw black, and the witness says they saw white.

Then, along the edges of the tree, write down the probabilities of each branch. For the top level, $P($actually black$) = 0.7$, $P($actually white$) = 0.3$. For each subcase of these two realities, the witness can either claim they saw black or white, and they are right 70% of the time, so for the black subcases $P($witness saw black$) = 0.7$ and $P($witness saw white$) = 0.3$, and for the white subcases $P($witness saw black$) = 0.3$ and $P($witness saw white$) = 0.7$.

The final step is to calculate the conditional probability (you're right that it involves the topic):

\begin{align}P(\text{actually white}\mid\text{witness saw white}) = & ~ \frac{P(\text{actually white and witness white}) }{ P(\text{witness white})} \\[1ex] = & ~ \frac{0.3 * 0.7 }{ (0.7 * 0.3 + 0.3 * 0.7)} \\[1ex] = & ~ 0.5 \end{align}

I get 50%. For the witness to say the car was white, there are two ways this could happen. The car was white and witness was correct $(.3)(.7)=.21$ and the car was black and the witness was incorrect $(.7)(.3)=.21$. These both happen with equal probability so there is a 50% chance the witness was correct. $$\frac{.21}{.21+.21}=.5$$