Easy probability (venn diagrams/conditional pr)

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The discussion focuses on solving probability problems using Venn diagrams and conditional probability. For the Venn diagram questions, the correct probabilities were calculated as Pr(A∩B) = 1/11, Pr(AUB) = 25/33, and Pr(AUB') = 6/11 after clarification. The second question involves finding the probability of drawing the ace of spades given that at least one ace is drawn, utilizing the conditional probability formula Pr(A|B) = Pr(A∩B)/Pr(B). The participant seeks guidance on calculating Pr(no aces) and Pr(A∩B) for the card problem. Overall, the thread emphasizes understanding and applying probability concepts accurately.
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Q1. Venn diagrams.

[PLAIN]http://img826.imageshack.us/img826/3194/23872542.jpg
Find:
a) Pr(A∩B)
b) Pr(AUB)
c) Pr(AUB')

a) this is easy, Pr(A∩B)=3/33=1/11
b) Pr(AUB)=(10+3+12)/33=25/33
c) this is the one I am not 100% sure on. I looked at the set of A and the set of B' seperately then took the intersection, I got Pr(AUB') = (10+3+8)/33 = 21/33

Q2. conditional pr.
Out of a standard deck of 52 cards, 5 cards are chosen at random. What is the probability that the ace of spades is chosen given at least one ace is chosen.

so I'm looking at conditional probability and the formula Pr(A|B) = Pr(A∩B)/Pr(B), where Pr(A) is probability of choosing ace of spades, and Pr(B) is probability of choosing at least one ace.

Pr(B) is simply 1-Pr(no aces).

Is this the right way to go about this question? If so, how do I find pr(no aces) and Pr(A∩B)?
 
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t_n_p said:
Q1. Venn diagrams.

[PLAIN]http://img826.imageshack.us/img826/3194/23872542.jpg
Find:
a) Pr(A∩B)
b) Pr(AUB)
c) Pr(AUB')

a) this is easy, Pr(A∩B)=3/33=1/11
b) Pr(AUB)=(10+3+12)/33=25/33
Good!

c) this is the one I am not 100% sure on. I looked at the set of A and the set of B' seperately then took the intersection, I got Pr(AUB') = (10+3+8)/33 = 21/33
No. [math]A\cup B'[/math] contains 10+ 8= 18 (10= things in A that are not in B, 8= things that are not in A and also not in B). [math]Pr(A\cup B')= 18/33= 6/11[/math].

Q2. conditional pr.
Out of a standard deck of 52 cards, 5 cards are chosen at random. What is the probability that the ace of spades is chosen given at least one ace is chosen.

so I'm looking at conditional probability and the formula Pr(A|B) = Pr(A∩B)/Pr(B), where Pr(A) is probability of choosing ace of spades, and Pr(B) is probability of choosing at least one ace.

Pr(B) is simply 1-Pr(no aces).

Is this the right way to go about this question? If so, how do I find pr(no aces) and Pr(A∩B)?
 
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Ah, so the 3 does not fall under A, but rather in its own group A∩B?
By the same token, then Pr (A'UB) = (12+8)/33?

What about the card problem, any ideas?
Thanks!
 
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