conditional probability trick questions - drawing cards from a deck and the meaning of 'at least'

Question

Suppose that a box contains one blue card and four red cards, which are labeled A, B, C, and D. Suppose also that two of these five cards are selected at random, without replacement.

a. If it is known that card A has been selected, what is the probability that both cards are red?

b. If it is known that at least one red card has been selected, what is the probability that both cards are red?

I have been assigned the problem above in a class. I am aware that the book considers the answer for A) to be 3/4, which is obvious. However, the answer to B) is claimed to be \begin{equation} P(red_1)P(red_2|red_1) = 4/5 \cdot 3/4 = 3/5 \end{equation}

The professor has not been able to explain why the answers to A and B are different in any way that makes sense to me. There is no way in which you cannot draw at least 1 red card in a draw of 2 cards. If you are told you have drawn at least 1 red card, then you have the same information that you did in A.

Answer 1

Look at the possible samples consistent with the information, each of which is equally likely:

$$0\color{red}{A}, 0\color{red}{B}, 0\color{red}{C}, 0\color{red}{D}, \quad \color{red}{AB},\color{red}{AC},\color{red}{AD},\color{red}{BC},\color{red}{BD},\color{red}{CD}$$

(The blue card is denoted "$0$".) The number of samples with both cards red is $6$; the total number of samples is $10$, whence the probability is

$$\Pr(\text{Both red}|\text{one red}) = 6/10 = 3/5.$$

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Contrast this situation with (a), where the possible samples are only $$0\color{red}{A}, \quad \color{red}{AB},\color{red}{AC},\color{red}{AD}$$

There, the chance that both cards are red is $3/4$.

Answer 2

For question A), the source of information is ambiguous. Do you have an A-Oracle who always tells you whether card A is among the two chosen cards, or do you have a Red-Oracle who always gives you the name of one of the red cards chosen?

The answer in the A-Oracle case is 3/4, as has been discussed.

The answer in the Red-Oracle case depends on how the Red-Oracle breaks ties. If, for example, the Red-Oracle always says A if there is an A, then he is just an A-Oracle in disguise, and the answer is 3/4. If, on the other hand, he chooses uniformly at random which red card in the two card hand to tell you about, then his answer is uninformative regarding whether there are one or two cards, so the answer is 3/5.

This may seem like nitpicking : this is probably a homework problem in the conditional probability section in a stats course, so the thing to do is to write out the simplest relevant probability space and cross out the entries that don't fit the information. On the other hand, the famous Monty Hall Problem is a conditional probability problem that confuses a lot of smart people, arguably because it is unclear what the source of information is.

Answer 3

Let $A_1$ - event such that first card is red, $A_2$ - event such that second card is red. Then we want to compute $P(\left( A_1 \cap A_2 \right) {|} \left(A_1 \cup A_2 \right))$, which is equal to $\frac{P(\left( A_1 \cap A_2 \right) \cap \left(A_1 \cup A_2 \right))}{P\left(A_1 \cup A_2 \right)} = \frac{P\left( A_1 \cap A_2 \right)}{P\left(A_1 \cup A_2 \right)}$. Let's notice that $P\left(A_1 \cup A_2 \right)$ is equal to 1, since there will be always at least one red card. Finally $P\left( A_1 \cap A_2 \right) = P(A_1)P(A_2 | A_1) = \frac{4}{5} \cdot \frac{3}{4} = \frac{3}{5}$. You can also prove that $P\left(A_1 \cup A_2 \right) = 1.$ $P\left(A_1 \cup A_2 \right) = \frac{\binom{4}{2} + \binom{4}{1}}{\binom{5}{2}}$