Question 656328: Hi, I really need help with this problem its been years since I took math 5 and this is driving me crazy.. can anyone explain step by step to me please? i will really appreciate it..
If i draw two cards at once (without replacement, find the following probabilities?
a) at least one face card
b) exactly one face card or one ace (but not one of each)
c) one spade and one heart
d) at least one face card or one club
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
At least one face. Difficult to calculate straight up, so go at the problem from a different direction. If you have drawn 2 cards, then there are exactly four possible situations that could occur, namely: 2 face, 1st one is a face and the second not, the 1st one is not a face and the second one is, and then no face cards at all. The sum of these four probabilities must be 1 because they are the only four things that can happen.
So, the easy way to calculate the probability of any of the first three situations occuring is to calculate the probability of the fourth situation and subtracting that value from 1.
The probability that the first card drawn is NOT a face card is the number of non-face cards in a deck divided by the number of cards in the deck, or  which reduces to  . Now, assuming that the first card drawn was a non-face, there are now a total of 39 non-face cards left in a deck that consists of 51 cards, hence the probability on the second draw is  . Since the two events as described are independent of each other, the probability of both is the product of their individual probabilities.
John
My calculator said it, I believe it, that settles it
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