- #1
SammC
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Homework Statement
In the following you are given a 5-card hand from a 52 card deck.
a) given that you have at least one ace, what is the probability you have at least 2 aces?
b) given that you have the ace of diamonds, what is the probability that you have another ace?
c) given that you have a red ace, what is the probability you have another ace?
d) Suppose you select a card from your hand at random and it is an ace. What is the probability you have another ace.
Homework Equations
pr(not a) = 1 - pr(a)
pr(A given B) = Pr(A and B) / Pr(B)
The Attempt at a Solution
C(X,Y) is X choose Y here.
Pr(A): The chance of drawing AT LEAST one Ace is 1 - C(48,5)/C(52,5) =~ .6588
Pr(B): The probability of having at least 2 aces: ?
Pr(A and B): ?
pr(C): That you have the ace of diamonds in a 5 card hand is 1-C(51,5)/C(52,5) =~ .096
Pr(C and B):?
pr(D): That you have at least one red ace in a 5 card hand: 1-C(50,5)/C(52,5) =~ .1848
(im not sure if i want the probability of having exactly 1 red ace, or at least one)
Pr(D and B):?
Finding all of the "?" probabilities should allow me to calculate problems a-c, but i really don't know how to find PR(X AND Y) with these problems, nor do I know how to get the probability of having at least two aces.