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- #1

#### nickar1172

##### New member

- Dec 11, 2013

- 20

a) solving for neither set A and neither set B

b) Solving for set A or Set B, but not both

c) Solving for set B given set A

d) Solving for set B given that he does not have Set A

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- Thread starter
- #1

- Dec 11, 2013

- 20

a) solving for neither set A and neither set B

b) Solving for set A or Set B, but not both

c) Solving for set B given set A

d) Solving for set B given that he does not have Set A

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- #2

- Jan 26, 2012

- 4,041

I probably need some more information. Are you simply asked to write these concepts mathematically, like $A \cap B$?

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- #3

- Dec 11, 2013

- 20

Yes that is exactly what I am talking about

I probably need some more information. Are you simply asked to write these concepts mathematically, like $A \cap B$?

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- #4

- Jan 26, 2012

- 4,041

Ok well we can use $A, A^c, B, B^c$ for referring to the sets and we can also use $\cup, \cap$. What do you get for the first one?Yes that is exactly what I am talking about

(by the way, $\cup$ means "or" and $\cap$ means "and")

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- #5

- Dec 11, 2013

- 20

Postulates for Probability - P(A) >= 0, P(S) = 1

Special Addition Rule (mutually exclusive) - If A ∩ B = ∅, then P(A ∪ B) = P(A) + P(B)

General Addition Rule - P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Probability and Odds (odds of A to B) - a / (a + b)

Conditional Probability - P(A | B) = P(A ∩ B) / P(B) provided that P(B) ≠ 0

General Multiplication Rule - P(A ∩ B) = P(A) . P(B | A)

Special Multiplication Rule(Independent events) - P(A ∩ B) = P(A) . P(B)

Multiplication Rules - if event B is independent of event A, then P(B | A) = P(A)

according to any of these rules how do you answer those question because I have NO IDEA how to

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- #6

- Jan 26, 2012

- 4,041

Ok, (a) is $A^c \cap B^c$. Put into words this is "not A and not B". For (b), what is "A or B" in symbolic language? What is "A and B"?

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- #7

- Dec 11, 2013

- 20