probability of an event cards We are given five cards. Three of the cards are black and they are numbered 1; 2; 3. The other two cards are red and they are numbered 1; 2. We pick two cards at random.
(a) What is the probability that both cards are red?
(b) What is the probability that both cards are red, if we know that at least one of them is red?
(c) What is the probability that both cards are red, if we know that one of them is red card number 1?
a) $1 / 10 = 0.10$
I don't really understand the difference between the question b and c. It should be the same, $1 / 4$ which is $0.25$ for both.
Am I right?
Thanks! 
 A: HINT:
Let $A$ and $B$ the two cards you get and we says $A=r$ if $A$ is red and $A=k$ if the card have the number $k$, then you have for b) (using the Bayes formula) that
$$\begin{align}P(A=&r\land B=r|A=r\lor B=r)=\frac{P(A=r\land B=r)}{P(A=r\lor B=r)}\\&=\frac{P(A=r\land B=r)}{P(A=r)+P(B=r)-P(A=r\land B=r)}\end{align}$$
And for c) you have
$$\begin{align}P(A=&r\land B=r|(A=r\land A=1)\lor(B=r\land B=1))\\&=\frac{P(A=r\land B=r)}{P((A=r\land A=1)\lor(B=r\land B=1))}\\&=\frac{P(A=r\land B=r)}{P(A=r\land A=1)+P(B=r\land B=1)}\end{align}$$
because the events $A=r\land A=1$ and $B=r\land B=1$ are mutually exclusive (that is, they cannot be at once).
By last to evaluate easily the individual probabilities choose an order of pick, say you pick card $A$ before card $B$, then $P(A=r)=3/5$ and (using the law of total probability) we have that
$$\begin{align}P(B=r)&=P(B=r|A=r)P(A=r)+P(B=r|A\neq r)P(A\neq r)\\&=\frac24\cdot\frac35+\frac34\cdot\frac25=\frac35\end{align}$$
as expected, because the probabilities for $A$ and $B$ are symmetric and hence must be equal. 
And for a), using again the law of total probability, we have that $$\begin{align}P(A&=r\land B=r)\\&=P(A=r\land B=r|A=r)P(A=r)+\overbrace{P(A=r\land B=r|A\neq r)}^{=\, 0}P(A\neq r)\\&=P(B=r|A=r)P(A=r)=\frac24\cdot\frac35=\frac3{10}\end{align}$$ 
And by last its easy to check that $P(A=r\land A=1)=P(B=r\land B=1)=1/5$.
