I am studying maths as a hobby and have come across this problem:
2 cards are drawn simultaneously from a pack of 52. What is the probability that both are spades if one card in known to be a) black, b) red, c) a spade, d) a king?
I've tried a few approaches.
I worked out all the things I thought might be relevant.
e.g. Total ways of drawing 2 spades is $\binom {13}{2} = 78$, total ways of drawing black cards = $\binom {26}{2} = 325$, total ways of drawing any 2 cards = $\binom {52}{2} = 1326$ but I cannot find how to use these to get the answer.
Then I tried to reason as follows: in c), if we know the first card is a spade the probability of drawing another spade is $\frac {12}{52}$
but none of these approaches lead me to the correct answers.
If we take $(c)$ for example,
The question says we know that one of the cards is a spade, what is the probability that both are spades? We need to apply conditional probability.
If Event $A$ is getting both as spades and Event $B$ is getting at least one spade in draw of $2$ cards then
i) $P(B) = \displaystyle \frac{{52 \choose 2} - {39 \choose 2}}{52 \choose 2} = \frac{15}{34}$
Numerator is number of ways to get at least one spade in a draw of $2$ cards, which is all possible draws of $2$ cards minus draws where none of the two cards is a spade.
ii) $ P(A\cap B) = \displaystyle \frac{{13 \choose 2}}{52 \choose 2} = \frac{1}{17}$
[please note that $P(A \cap B) = P(A)$]
So the desired probability $ \displaystyle P (A|B) = \frac{P(A\cap B)}{P(B)} = \frac{2}{15}$.
Now can you use this to evaluate others? If you get stuck, let me know.
While the above can be simplified a bit more, I have gone ahead and used standard working which is important to get used to for conditional probability.