Problem on Conditional Probability (from textbook by Sadler & Thorning) Q: A bag contains five discs, three of which are red. A box contains six discs, four of which are red. A card is selected at random from a normal pack of 52 cards, if the card is a club a disc is removed from the bag and if the card is not a club a disc is removed from the box.
Find:
(a) the probability that the disc removed will be red
(b) the probability that, if the removed disc is red it came from the bag.

The above is an example from Understanding Pure Mathematics by Sadler and Thorning. I think the given solution to (b) might be incorrect. Can someone please explain.

R = Red, C = Club
(a)
P(R) = P(C ∩ R) + P( C' ∩ R)
     = 3/20 + 1/2
     = 13/20

The part below I do not understand.
(b) We now require P(C|R)
P(C|R) = P(R ∩ C)/P(R)
   = (3/20)/(13/20)
   = 3/13


My question is, since the disc can only be picked AFTER the card is picked, shouldn't we find P(R|C)?
 A: Conditional probability works regardless of chronological order. The probability $P(C\mid R)$ answers the question: Having drawn a red disc, what's the probability that it came from the bag? This is the question that was asked and it appears to have been solved correctly.
Note that this has a different meaning than $P(R\mid C)$, which is the probability that we get a red disc while choosing from the bag.
A: Probability is typically about information, i.e. what we know and what we don't know.
So, it will help to read a conditional probability $P(A|B)$ as: "Given that you know that event $B$ has occurred, what is the probability that event $A$ has occurred?"
That makes it clear that the 'condition' in 'conditional probabilities' has nothing to do with the order of the events, but rather with what you know.
A: It refers to the posterior probability and the Bayesian theorem:
$$P(C|R)=\frac{P(R\cap C)}{P(R)}=\frac{P(R|C)\cdot P(C)}{P(C\cap R)+P(C'\cap R)}=$$
$$\frac{\frac{3}{5}\cdot \frac{13}{52}}{\frac{3}{20}+\frac{1}{2}}=\frac{\frac{3}{20}}{\frac{13}{20}}=\frac{3}{13}.$$
