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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?

  • 1
    It's curious how much this reminds me o$f$ [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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    @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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    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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    Can you help-me ?2012-05-04