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Possible Duplicate:
A probability problem

Let $A$ and $B$ be events, $P(A) = \frac{1}{4} $, $P(A\cup B) = \frac{1}{3} $ and $ P (B) = p $.

  1. Find $p$, if $A$ and $B$ are mutually exclusive.
  2. Find $p$, if $A$ and $B$ are independent.
  3. Find $p$, if $A$ is a subset $B$.

I know for 1) mutually exclusive: $P(A) + P(B) = P(A \cup B)$, but how I can find p ? I don't know how to solve it. Please help me.

Obs: Sorry for duplicate post.

  • 0
    Just substitute $$\underbrace{P(A)}_{=1/4}+\underbrace{P(B)}_{= p}=\underbrace{P(A\cup B)}_{=1/3}$$so$${1\over 4}+p={1\over3};$$now solve for $p$.2012-05-04
  • 0
    Ok, I find this result in 1.) $\frac{1}{12}$. But in the second.2012-05-04
  • 0
    Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$.2012-05-04
  • 0
    2) I know the independent events are: $P(A) * P(B) = P(A \cap B)$.2012-05-04
  • 2
    [This](http://math.stackexchange.com/questions/140836/a-probability-problem) question by the same user asks the same question. I don't see a need to start to a new post. One of them needs closure. Which one should be closed shall be decided by the community and I'll vote accordingly.2012-05-04

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