Probability to get a card in a card game Yesterday, while playing a card game, we had an argument about probabilities.
The game settings (for 4 players): there are 2 red cards ($R$) and 3 blue cards ($B$). At the beginning of a round, those 5 cards are shuffled and one is taken out. Then the card are distributed one by one to the 4 players.


*

*What is the probability that 2 red card are present?

*What is the probability that only 1 red card is present?

*What is the probability that only 1 red card is present knowing that I have a blue card?


$P(2R)$
For this one, there is $\frac{2}{5}$ chances that a red card has been taken out and $\frac{3}{5}$ that a blue card has been taken out. $P(2R) = \frac{3}{5}$.
$P(1R) = \frac{2}{5}$ in a similar way.
This problem can also be considered the other way around. We got a set of 5 cards, and we pick them one by one until we have a set of 4, thus leaving one out.
For the first card, you have a $\frac{2}{5}$ chances to pick a red card. For the second card, you have a $\frac{1}{4}$ chances to pick a red card. Thus, you have a $\frac{1}{10}$ chance to pick 2 reds with the 2 first card.
If you do that for all combinations (with a tree to avoid mistakes), you get $\frac{1}{10}$ for each of the 6 branch ending with 2 reds. Thus, in total, you have $\frac{6}{10} = \frac{3}{5}$ to have 2 reds in play.
Next the conditional probability: $P_B(1R) = \frac{P(B\bigcap1R)}{P(B)}$.
$P(B)$ is the probability that I get a blue card. Thus, I think it is 
$P(B) = \frac{3}{4}*P(3B, 1R) + \frac{2}{4}*P(2B, 2R) = \frac{3}{4}*\frac{2}{5} + \frac{2}{4}*\frac{3}{5} = \frac{3}{5}$.
Next, and that's the one I am unsure about: $P(B\bigcap1R) = P(B)*P(1R) = \frac{3}{5}*\frac{2}{5} = \frac{6}{25}$.
Why am I unsure? If I recall correctly, this multiplication can be done if both events are independent. Can we say that those 2 events are independent? What if they are not?
And thus: $P_B(1R) = \frac{6}{15}$
How wrong am I? Thanks for the help :)


What is the probability that two red cards are present?

As you observed, the probability that two red cards are present is equal to the probability that a blue card was removed, which is 
$$\Pr(\text{two red cards are present}) = \Pr(\text{a blue card is removed}) = \frac{3}{5}$$
as you found
As for your failed attempt, observe that there are $\binom{5}{2}$ ways to pick two positions for the red cards.  Two red cards can be found in the first four positions in $\binom{4}{2}$ ways.  Hence, the probability that two red cards are in the first four positions is
$$\Pr(\text{two red cards are present}) = \frac{\dbinom{4}{2}}{\dbinom{5}{2}} = \frac{6}{10} = \frac{3}{5}$$
What was your mistake?
You did not take into account the probability of selecting a blue ball?
Your first calculation is correct.  The probability that red balls are in positions 1 and 2 is 
$$\Pr(\text{red, red}) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$$
When you calculated the probability that red balls are in positions 1 and 3, you did not take into account the probability that a blue ball was picked second.
$$\Pr(\text{red, blue, red}) = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{10}$$
The fact that both calculations yielded $\frac{1}{10}$ is not a coincidence.  Let's do those two calculations again, this time writing out all five terms.
\begin{align*}
\Pr(\text{red, red, blue, blue, blue} & = \frac{2}{5} \cdot \frac{1}{4} \cdot \frac{3}{3} \cdot \frac{2}{2} \cdot \frac{1}{1} = \frac{3!2!}{5!} = \frac{1}{10}\\
\Pr(\text{red, blue, red, blue, blue} & = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \cdot \frac{2}{2} \cdot \frac{1}{1} = \frac{3!2!}{5!} = \frac{1}{10}
\end{align*}
In fact, the probability that the two red balls appear in any two positions of the sequence is always $1/10$ since there are $\binom{5}{2} = 10$ equally likely ways to place the red balls in the sequence of two red and three blue balls.

What is the probability that only one red card is present?

As you observed, the probability that only one red card is present is equal to the probability that a red card is removed, which is 
$$\Pr(\text{only one red card is present}) = \Pr(\text{a red card is removed}) = \frac{2}{5}$$
as you found.

What is the probability that only one red card is present given that you have a blue card?

Since you are equally likely to have any of the cards, the probability that you have a blue card is $3/5$.  
Clearly, your blue card could not have been removed.  Therefore, the card that is removed must be one of the other four cards, of which two are red.  Since a red card must be removed if only one red card is present, the probability that you only one red card is present given that you have a blue card is 
$$\Pr(\text{only one red card is present} \mid \text{you have a blue card}) = \frac{2}{4} = \frac{1}{2}$$ 
which is the answer to your question.
What was your mistake?
If we denote the event only one card is present by $1R$ and the event that you have a blue card by $B$, then the probability that you have a blue card and only one red card is present is 
$$\Pr(1R \cap B) = \Pr(B)\Pr(1R \mid B) = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}$$
How could you have seen this directly?  
If you write the sequences of three blue and two red cards in which a blue card is in the first position (representing the card you hold), there are three ways for a red card to be in the fifth position, namely those in which the other red card is in the second, third, or fourth positions. Hence, three of the ten sequences of three blue and two red cards have you holding a blue card with only red card present in the game.
