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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
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- #1

- Jan 26, 2012

- 4,055

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Jan 26, 2012

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Congratulations to the following members for their correct solutions:

1) MarkFL

Solution:
__Remark__: I want to comment on an error that was made by a few submissions. For this problem and all similar problems it is understood that there is a sample space, $S$, that represents all possible outcomes of a random experiment. $A$ and $B$ are both subsets of $S$ and can be called events, but aren't necessarily the only two events unless stated so. Two important axioms that are associated with these definitions are (1) $\forall E, 0 \le P(E) \le 1$ and (2) $P(S)=1$, where $E$ is any possible event. Everyone remembered (1) quite well but forgot about (2). (2) implies that if $P(A)+P(B)>1$ then it follows that $A$ and $B$ are not disjoint.

[HR][/HR]

We are asked to maximize \(\displaystyle P(A \cup B) - P(A \cap B)\).

Starting with the formula for adding the probabilities of two events which may or may not be mutually exclusive we have $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. We can subtract $P(A \cap B)$ from both sides of this identity to find another useful identity: \(\displaystyle P(A \cup B) - P(A \cap B)=P(A)+P(B)-2P(A\cap B)\).

$P(A)$ and $P(B)$ are given to us and can't change, so we notice that the way to maximize $P(A)+P(B)-2P(A\cap B)$ is to minimize $2P(A\cap B)$. To make this easier to do, we can first minimize $P(A\cap B)$, which in turn minimizes $2P(A\cap B)$.

Once again uses the identity $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and plugging in known values we get $P(A\cup B)=1.6-P(A\cap B)$. Lastly we note that $P(A \cup B) \le 1$. If it's exactly 1 that means that $A$ and $B$ contain all possible outcomes in $S$. If it's less than 1 then there are some outcomes outside of $A$ and $B$ but still in $S$. To minimize $P(A\cap B)$ we choose the case when that union is 1 so we finally reach the conclusion that $1=1.6-P(A\cap B)$ or $P(A\cap B)=.6$.

Going back to this equation we found above, \(\displaystyle P(A \cup B) - P(A \cap B)=P(A)+P(B)-2P(A\cap B)\), and plugging in all known values we find the solution is 0.4.

1) MarkFL

Solution:

[HR][/HR]

We are asked to maximize \(\displaystyle P(A \cup B) - P(A \cap B)\).

Starting with the formula for adding the probabilities of two events which may or may not be mutually exclusive we have $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. We can subtract $P(A \cap B)$ from both sides of this identity to find another useful identity: \(\displaystyle P(A \cup B) - P(A \cap B)=P(A)+P(B)-2P(A\cap B)\).

$P(A)$ and $P(B)$ are given to us and can't change, so we notice that the way to maximize $P(A)+P(B)-2P(A\cap B)$ is to minimize $2P(A\cap B)$. To make this easier to do, we can first minimize $P(A\cap B)$, which in turn minimizes $2P(A\cap B)$.

Once again uses the identity $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and plugging in known values we get $P(A\cup B)=1.6-P(A\cap B)$. Lastly we note that $P(A \cup B) \le 1$. If it's exactly 1 that means that $A$ and $B$ contain all possible outcomes in $S$. If it's less than 1 then there are some outcomes outside of $A$ and $B$ but still in $S$. To minimize $P(A\cap B)$ we choose the case when that union is 1 so we finally reach the conclusion that $1=1.6-P(A\cap B)$ or $P(A\cap B)=.6$.

Going back to this equation we found above, \(\displaystyle P(A \cup B) - P(A \cap B)=P(A)+P(B)-2P(A\cap B)\), and plugging in all known values we find the solution is 0.4.

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