Stats probability addition rule, multination rule The directions are to calculate the following probability based on drawing cards without replacement from a standard deck of 52.
What is the probability of drawing a 2 or a king on the first draw and drawing a queen on the second draw? 
 A: So the Probability of drawing a $2$ or a King is the probability of drawing a $2$ + probability of drawing a King: $$P(2)= {4\over 52}={1\over 13}$$$$P(\text{King})={4\over 52}={1\over 13}$$ So the $P(2\cup \text{King})$ is:$$P(2\cup \text{King})={1\over 13}+{1\over 13}={2\over 13}$$ The probability of drawing a queen is just: $$P(\text{Queen})={1\over 13},$$ but since it is out of $51$ cards instead of $52$ you get $$P(\text{Queen*})={4\over51}$$ So now we can do $$P(A\cap B)=P(A)P(B) \qquad \qquad \text{Since they are independent}$$where $P(A)=P(2\cup \text{King})$ and $P(B)=P(\text{Queen*})$ 
So substituting and multiplying gives us $${2\over13}\cdot {4\over 51}={8\over 663}=.01206637$$
A: You have 52 cards, 4 of which are kings, 4 of which are $2$'s, 4 of which are queens.
The first event, 'drawing a $2$ or a king' should be the first thing considered. There are 8 cards which are a king OR a $2$, so there is a likelihood of $8/52$.
The second event, 'drawing a queen' is even simpler (when considered by itself). There are 4 queens, 52 cards... so assuming a full deck, $1/13$ is your odds of a queen.
This is not quite the question though! You are assuming that you have already removed a card on the first draw, so you have the probability of success on the first trial: $8/52$ and the odds likelihood of a queen is 4 out of now 51 cards (or $4/51$).
The likilihood of BOTH events is thus: $(8/52)*(4/51)$ which is approximately $1.2$%
