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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.

  • 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

2 Answers 2

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(1) You know that if $A$ and $B$ are mutually exclusive, then $P(A\cup B)=P(A)+P(B)$. Using the values given to you, you have $\frac13=\frac14+p$; just solve this for $p$ by subtracting $\frac14$ from both sides of the equation.

(2) This one is a bit harder. $A$ and $B$ are independent if $P(A\cap B)=P(A)\cdot P(B)$, so we want to solve $\frac14p=P(A\cap B)$. Unfortunately, we aren't given $P(A\cap B)$, so we have to find it. Use the fact that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, so that $P(A\cap B)=P(A)+P(B)-P(A\cup B)$. In your problem this becomes $P(A\cap B)=\frac14+p-\frac13$. Moreover, since $A$ and $B$ are independent, this is all equal to $P(A)\cdot P(B)=\frac14p$. Thus, you simply need to solve the equation $\frac14p=\frac14+p-\frac13$ for $p$.

(3) This is easy: if $A$ is a subset of $B$, then $A\cup B=B$.

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    Thank you man, you made ​​me understand. – 2012-05-04
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You should have a law $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

For 2, you should have one that says for independent events $P(A \cap B)=P(A)P(B)$

For 3, if $A$ is a subset of $B$, then $P(A \cap B)=P(A), P(A \cup B)=P(B)$. Do you see why?