Probability that a cow is black given that I've observed at least one side is black

Question

I'm on a farm with six cows; three are white, two are black and one is completely black on one side and completely white on the other. I see one cow from the side, who appears to be black (that is, the side that I see is black). What's the probability that the cow is black?

I tried to figure out a solution, but it's wrong as it conflicts with the solution I'm given. I don't understand completely what I'm missing and where I'm going wrong, so I would appreciate some advice which would lead me in the right direction.

Let $P$ be the event that the cow I observe is completely black, and $S$ be the event that at least one side is black. $S$ is given, so I need to find $P(B|S)$.

$P(B|S) = \frac{P(S|B)P(B)}{P(S)}$

$P(S|B) = 1$, since if the entire cow is black then at least one side is black. $P(B) = \frac{1}{3}$ since there are two black cows among a group of six. And $P(S) = \frac{1}{2}$ since there are three out of six cows with at least one black side.

so $P(B|S) = \frac{P(B)}{P(S)} = \frac{2}{3}$

However the answer I'm given is $\frac{4}{5}$

Answer 1

Do you mean your first sentence to read "three are white and two are black"? There's a contradiction between that sentence and your computation of $P(B)$.

Also, to compute $P(S)$, I think you want to count how many black sides there are among all cows, not the number of cows with at least one black side. You're equally likely to see any side of any cow.

Answer 2

$P(S)$ is not $1/2$. There are $12$ sides you can see, all equally likely, and $5$ of the sides are black.

So $P(S) = 5/12$. This yields $P(B \mid S) = (1/3)/(5/12) = 4/5$.

Answer 3

Sometime probability can be counterintuitive.

The key here is to note that "you see the black side of a cow" is a different event from "you see a cow with a black side".   (Consider that if the two coloured cow were to walk passed with the white side towards you, then you would not see the black side of a cow but you would see a cow with a black side.)

You are four times as likely to see the black side of one of the all black cows as you are to see the black side of the two coloured cow. (There are two of them, and they each have two black sides.)

So $S$, the event that at least one side is black, is the wrong event to use. You need $B$, the event that the side which is seen is black.

So the probability that the cow is all black when given that you see a black side of the cow is: $$\mathsf P(A\mid B) = \frac{\mathsf P(A\cap B)}{\mathsf P(T\cap B)+\mathsf P(A\cap B)} = \frac{4/12}{1/12+4/12} = \frac 4 5$$