Problem on Conditional Probability (from textbook by Sadler & Thorning)

Question

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.

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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.

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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

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My question is, since the disc can only be picked AFTER the card is picked, shouldn't we find P(R|C)?

Answer 1

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.

Answer 2

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.

Answer 3

It refers to the posterior probability and the Bayesian theorem: $$P(C|R)=\frac{P(R\cap C)}{P(R)}=\frac{P(R|C)\cdot P(C)}{P(C\cap R)+P(C'\cap R)}=$$ $$\frac{\frac{3}{5}\cdot \frac{13}{52}}{\frac{3}{20}+\frac{1}{2}}=\frac{\frac{3}{20}}{\frac{13}{20}}=\frac{3}{13}.$$