Let A,B be two events such that P(A) = .4 ; P(A U B)= .7. Let P(B)= p

Oct 2013
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Let A,B be two events such that P(A) = .4 ; P(A U B)= .7. Let P(B)= p

a.) For what choice of p are A and B mutually exclusive events?

b.) For what choice of p are A and B independent events?

I know P(A U B) = P(A) + P(B) - P(A ∩ B) thus P(A ∩ B) = 0.4 + P(B)-0.7 but I don't know how I would find P(A ∩ B) without knowing P(B)
 

romsek

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Nov 2013
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draw some venn diagrams and think it out from first principles

for (b)

if two events are independent what is $P(A \cap B)$ ?

and you do know P(B), it's given as p
 
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Plato

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Aug 2006
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Let A,B be two events such that P(A) = .4 ; P(A U B)= .7. Let P(B)= p
a.) For what choice of p are A and B mutually exclusive events?
If two events are mutually exclusive then $A\cap B=\emptyset$ so $\mathcal{P}(B)=?$

Let A,B be two events such that P(A) = .4 ; P(A U B)= .7. Let P(B)= p
b.) For what choice of p are A and B independent events?
If two events are independent then $\mathcal{P}(A\cap B)=\mathcal{P}(A)\mathcal{P}(B)$.

So solve this $0.7=0.4+\mathcal{P}(B)-0.4\cdot\mathcal{P}(B).$
 
Last edited:
Oct 2013
46
0
USA
So part a.) would be p= 0.3 and part b.) would be 0.5 correct? I'm a bit confused as to where the formula used to solve b came from
 

romsek

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Nov 2013
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So part a.) would be p= 0.3 and part b.) would be 0.5 correct? I'm a bit confused as to where the formula used to solve b came from
a) is correct

$P[A \cup B]=P[A] + P - P[A \cap B]$
 
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SlipEternal

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Nov 2010
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So part a.) would be p= 0.3 and part b.) would be 0.5 correct? I'm a bit confused as to where the formula used to solve b came from
It came from your formula. Instead of $P(A\cap B)$, Plato replaced it with $P(A)P(B) = 0.4P(B)$.
 
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