# 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.

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$) = 0.7$.
\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. $$\frac{.21}{.21+.21}=.5$$