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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 $B$.

Can someone help me to solve it?

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    Though there are hints now, what have you tried? Please tag the home work questions as such; don't worry--you'll get well thought out hints that should help you complete the solution. Regards,2012-05-04
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    It's curious how much this reminds me of [this problem](http://math.stackexchange.com/questions/12941/nice-puzzle-100-bread-rings-and-two-bags) where several people agreed that the solution was "not mathematical".2012-05-04

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Hint.1: $P(A \cup B) = P(A) + P (B)$ when the two events are mutually exclusive.

Hint.2: $P(A \cap B) = P(A) \cdot P (B)$ when the two events are independent.

Hint.3: $P(A \cap B) = P(A) $ when $A$ is a subset of $B$.

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    So, ok, I find this result in I) 1 = $P(A) + P(B)$, SO ... 1 = $\frac{1}{4} + $p SO ... p = $\frac{1}{1} - \frac{1}{4} = \frac{3}{4}$2012-05-04
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    That is right ?2012-05-04
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    That's right @mastergoo. Well done.2012-05-04
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    II) $1 = P(A) * P(B)$ ... ... $1 = \frac{1}{3} * p$ ... ... $p = \frac{1}{\frac{1}{3}}$ ... ... $p = 3$2012-05-04
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    II) That's right ?2012-05-04
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    @mastergoo: No, that's wrong. What you have is $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ and $P(A \cap B) = P(A) \cdot P (B)$.2012-05-04
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    I suspect the other one was wrong as well, you need to be more careful.2012-05-04
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    So I don't know how to solve this question, please be more specific.2012-05-04
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    In the question I) I've to use the P(A)+P(B)−P(A∩B) OR P(A)+P(B) ?2012-05-04
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    And if I tell that the answer of the exercice I) is $\frac{1}{12}$, is right ?2012-05-04
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    @mastergoo: 1) $1/3=1/4+p$ , 2) $1/3=1/4+p+p/4$ , 3) ... _ The main formula is $P(A \cup B)=P(A)+p(B)−P(A \cap B)$ and you substitute $P(A \cap B)$ with what I said in each part!2012-05-04
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  • $ A$ and $B$ are mutually exclusive: $P(A \cap B)= 0$
    $P(A \cup B) = P(A) + P(B)$
    Then you can find value of $p$ using it.

  • $A$ and $B$ are independent: $P(A \cap B)= P(A) \cdot P(B)$
    $P(A \cup B) = P(A) + P(B)-P(A) \cdot P(B)$
    Then you can find value of p using it.

  • $A$ is a subset of $B$, that is, $A \subseteq B$: $A \cup B = B$
    That implies $P(A \cup B) = P(B)$
    Then you can find value of p using it.

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    Dear Prasad, +1; Happy to see you contributing. I wanted to let you know that the site allows TeX mark up for the Math. Please TeXify your posts to add to your well-thought out answers. Regards,2012-05-04
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    Thank you for your suggestion and i am learning2012-05-04
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    How can I solve III) ?2012-05-04
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    Can you help-me ?2012-05-04
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If $A$ and $B$ are independent events then $p(A \cap B) = p(A) \cdot p(B)$ and note that

  • $p(A \cup B) = p(A) + p(B) - p(A \cap B)$

This answers ii

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    How can I solve III) ?2012-05-04
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    @mastergoo: See the first answer2012-05-04
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    Ok, I saw but I can not understand. Logicaly if A is subset of B, A U B = B.2012-05-04
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    @mastergoo: what does $A \subset B$ mean, it means all elements in $A$ are in $B$. Now, from this you have$A \cap B = A$, and hence $p(A\cap B) = p(A)$2012-05-04
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    Oh, OK, I understand now.. My solution for I and II are right ?2012-05-04
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    Can you help-me ?2012-05-04