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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.

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marked as duplicate by Austin Mohr, Did, Pedro Tamaroff, t.b., Asaf Karagila Jun 2 '12 at 21:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 $p$. $\endgroup$ – David Mitra May 4 '12 at 17:47
  • $\begingroup$ Ok, I find this result in 1.) $\frac{1}{12}$. But in the second. $\endgroup$ – mastergoo May 4 '12 at 17:48
  • $\begingroup$ Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$. $\endgroup$ – David Mitra May 4 '12 at 17:50
  • $\begingroup$ 2) I know the independent events are: $P(A) * P(B) = P(A \cap B)$. $\endgroup$ – mastergoo May 4 '12 at 17:51
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    $\begingroup$ This 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. $\endgroup$ – user21436 May 4 '12 at 17:59
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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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  • $\begingroup$ in 3)..... P (B) = p, but I've to find p ? $\endgroup$ – mastergoo May 4 '12 at 18:04
  • $\begingroup$ @mastergoo: But you know $P(A\cup B)$, which is the same thing. $\endgroup$ – Brian M. Scott May 4 '12 at 18:05
  • $\begingroup$ I'm confused yet.. I've to find p if A is subset of B, if A inside B, how I can find p ????????? $\endgroup$ – mastergoo May 4 '12 at 18:06
  • $\begingroup$ @mastergoo: In (3), $A\cup B$ and $B$ are the same event: $A$ is already part of $B$, so taking the union with $A$ adds nothing to $B$. Thus, $A\cup B$ and $B$ have the same probability. You know the probability of $A\cup B$, so you know the probability of $B$. It's ... ? $\endgroup$ – Brian M. Scott May 4 '12 at 18:08
  • $\begingroup$ Wow I understood.. To find p, A inside B, the union is B. In other words $ P(A \cup B) = \frac{1}{3}$. $\endgroup$ – mastergoo May 4 '12 at 18:09
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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?

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