Bayes theorem, 3 cards problem

Question

i have to different solutions from my professor to the same problem (so there must be a mistake somewhere).

The problem is as follows, we have 3 cards, - one is blue on both sides - one is red on both sides - one is blue on one and red on the other side

If a card is drawn randomly, what is the probability that given one side red, the other side is blue.

One of the answers is 1/2, which seems logical to me as there are 2 blue cards, one is all blue and one is blue and red. So the prob. That the other side is red is 1/2.

The other answer is 1/3, and i have actually no clue how he derives there, we got sent a paper with all answers to problems we did in class....

When using Bayes theorem I have a problem defining events, i tried as follows

P(RB|R)=P(RB)*P(R|RB)/P(R)

Where r is a card with at least one red side.

=(1/3*1)/(2/3)=1/2

Many thanks for your help

Answer 1

Technically, you don't have to use Bayes's rule here. I think it's more "immediate" to use the conditional probability formula (see more here and here).

So let's solve your question. The following illustration may be of help (taken from here):

You asked:

what is the probability that given one side red, the other side is blue

Apply to conditional probability formula:

$$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} $$

$$ P(\text{red-blue card} \mid \text{red side}) = \frac{P(\text{red-blue card} \cap \text{red side})}{P(\text{red side})}$$

Ok, so what's ${P(\text{red-blue card} \cap \text{red side})}$?

By definition, that's the probability that both events occur, meaning, the probability we get a bi-color card, and then a red side. There's only a single bi-color card, and the probability of selecting it is $\frac{1}{3}$. Picking the red side has a probability of $\frac{1}{2}$, so the probability of their intersection is $\frac{1}{6}$.

Now, what's ${P(\text{red side})}$?

That's the probability of drawing a red side. We have 3 red sides out of 6 sides, so the probability of drawing a red side is $\frac{1}{2}$.

In summary:

$$ P(\text{red-blue card} \mid \text{red side}) = \frac{P(\text{red-blue card} \cap \text{red side})}{P(\text{red side})} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$

Answer 2

It's just a counting problem. Take the number of ways you can have blue on both sides and divide it by the number of ways you can have blue on the side you look at.

The trick is, we are counting sides, not cards. There are three blue sides, and hence three ways to be "given" a blue side. But there is only one way to have both sides be blue. So $P(\text{back is blue}|\text{front is blue})=1/3$.