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Card Hand probabilities: Discrete Math

  • Thread starter SammC
  • Start date
17
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1. 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.


2. Homework Equations
pr(not a) = 1 - pr(a)
pr(A given B) = Pr(A and B) / Pr(B)


3. 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 dont 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.
 

jambaugh

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Probability of having at least two aces...
Hmmm...
You can calculate the probability of having exactly 4 aces + exactly 3 aces + exactly 2 aces.

Pr(4) = C(4,4)C(48,1)/C(52,5)
Pr(3) = C(4,3)C(48,2)/C(52,5)
Pr(2) = C(4,2)C(48,3)/C(52,5)
Add 'em up.
 
17
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Thanks, that makes sense.

How do you go about getting the probability of the union though?
 

jambaugh

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Since you are enumerating distinct hands (no hand with exactly 2 aces is a hand with exactly one ace) you can simply add the probabilities.
Pr(x>1)= Pr(x=2) + Pr(x=3) + Pr(x=4).
 
17
0
what about pr(1 ace red AND at least 2 aces) one is not necessarily in the other in this case, same with 1 ace diamond
 

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