Easy probability (venn diagrams/conditional pr)

[t_n_p]

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)?

 

[HallsofIvy]

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. \(\displaystyle A\cup B'\) 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). \(\displaystyle Pr(A\cup B')= 18/33= 6/11\).

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)?

 

[t_n_p]

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!