Easy probability (venn diagrams/conditional pr)

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SUMMARY

This discussion focuses on solving probability problems using Venn diagrams and conditional probability. The first question involves calculating probabilities using the formulae Pr(A∩B), Pr(AUB), and Pr(AUB'). The correct answers are Pr(A∩B) = 1/11, Pr(AUB) = 25/33, and Pr(AUB') = 6/11. The second question addresses conditional probability, specifically Pr(A|B) = Pr(A∩B)/Pr(B), with Pr(B) calculated as 1 - Pr(no aces) when drawing from a standard deck of 52 cards.

PREREQUISITES
  • Understanding of Venn diagrams and their application in probability.
  • Familiarity with basic probability concepts, including intersection and union of sets.
  • Knowledge of conditional probability and its formula, Pr(A|B) = Pr(A∩B)/Pr(B).
  • Ability to calculate probabilities from a standard deck of cards.
NEXT STEPS
  • Study the principles of Venn diagrams in probability theory.
  • Learn how to calculate conditional probabilities in various scenarios.
  • Explore advanced probability concepts, such as Bayes' theorem.
  • Practice solving probability problems using real-world examples, such as card games.
USEFUL FOR

Students, educators, and professionals in mathematics or statistics who seek to enhance their understanding of probability concepts, particularly in relation to Venn diagrams and conditional probability.

t_n_p
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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' separately 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)?
 
Last edited by a moderator:
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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' separately 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)?
 
Last edited by a moderator:
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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