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

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Homework Help Overview

The problem involves calculating probabilities for independent events A and B, where P(A) = 0.30 and P(B) = 0.50. The specific tasks are to compute P(A U B) and P(A' ∩ B).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial attempts at calculating P(A U B) and P(A' ∩ B), noting that the original calculations appear incorrect. Questions arise about the meaning of independence and when the union of two events equals the sum of their probabilities. Some suggest using a Venn diagram for better understanding.

Discussion Status

There is an ongoing exploration of the concepts of independence and exclusivity, with some participants providing clarifications and corrections to the initial attempts. Guidance is offered regarding the correct formula for calculating the union of independent events, and there is recognition of the need to understand the implications of independence on the intersection of events.

Contextual Notes

Participants express uncertainty about the definitions and calculations involved, particularly regarding the intersection of independent events and the implications of the events not being exclusive.

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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[itex]\cap[/itex]B)
Look after how to calculate the probability P(A[itex]\cap[/itex]B) 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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