Question. So two cards are drawn without replacements from a pack of playing cards, I've been told to work out the Probability:
a) Both cards are aces b) Only one card is an ace c) Two cards are of different suits
Then I've been told Given at least one ace is drawn, Find the probability that the two cards are different suits?
Attempt: Now I have solved a) to c), for the last part I'm not getting the same answer as the book.
Let the events A = Two cards drawn of different suits & B = At least one ace drawn.
$P(A\mid B) = {P(A\cap B)}/{P(B)}$
$P(B)= a)+ b)=33/221$
$P(A\cap B)$=....this is where I'm stuck, I keep getting different values by considering the case 1 ace is picked then adding the case where 2 aces are picked.
Any help will be appreciated, Thanks.