Calculating Probabilities for Independent Events: A Union and A Intersection B

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Homework Statement


A and B are independent, P(A) = 0.30, P(B) = 0.50. Compute P(A U B) and P(A' intersect B).



The Attempt at a Solution


P(A U B) = 0.30 + 0.50 = 0.80
P(A' intersect B) = 1 - 0.30 = 0.70
this is obviously wrong
 
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vanitymdl said:

Homework Statement


A and B are independent, P(A) = 0.30, P(B) = 0.50. Compute P(A U B) and P(A' intersect B).



The Attempt at a Solution


P(A U B) = 0.30 + 0.50 = 0.80
P(A' intersect B) = 1 - 0.30 = 0.70
this is obviously wrong

It is really wrong. What does it mean that A and B are independent? When is it true that P(A U B) = P(A)+P(B)?
Try to draw the Venn diagram.

ehild
 
SammyS said:
The first one looks OK .

Yes, incorrect.

What is P(A') ?

Isn't P(A ∩ B) = 0 ? In other words, A ∩ B = ∅ , the null set.

So what is A' ∩ B and thus P(A' ∩ B) ?

Well I got the first one incorrect and I don't know exactly why because wouldn't the union be what is included in the set A and B. That's why I added them.

For the second on I'm no sure how to really do that one. I just understand it as not in A and in B
 
ehild said:
Independent does not mean exclusive. The two events can happen together. If A is throwing "6" with one dice and throwing "5" with the other dice, these events are independent but not exclusive.
http://www.mathgoodies.com/lessons/vol6/addition_rules.html

ehild
Yes, of course you're correct, ehild !

(I have deleted my earlier post. )
 
vanitymdl said:
Well I got the first one incorrect and I don't know exactly why because wouldn't the union be what is included in the set A and B. That's why I added them.

P(AUB)=P(A)+P(B)-P(A\capB)
Look after how to calculate the probability P(A\capB) if A and B are independent.

vanitymdl said:
For the second on I'm no sure how to really do that one. I just understand it as not in A and in B

It is the probability that A does not happen and B happens.

ehild
 
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