# Probability Example Question - Understanding Independence

An insurance company pays hospital claims. The number of claims that include emergency room or operating room charges is 85% of the total number of claims. The number of claims that do not include emergency room charges is 25% of the total number of claims. The occurrence of emergency room charges is independent of the occurrence of operating room charges or hospital claims. Calculate the probability that a claim to the insurance company includes operating room charges.

From the question stem, I interpreted the given data as follows:

Let E = Emergency room charges, and O= operating room charges

$\Pr(E \cup O) = 0.85$

$\Pr(E') = 0.25$

$\Pr(E) = 0.75$

The part of the question stem that confused me is this statement:

The occurrence of emergency room charges is independent of the occurrence of operating room charges or hospital claims.

Why is it necessary to say or hospital claims? At first, I understood that there were two sets, namely Emergency room charges and Operating room charges, but when they state or hospital claims, it makes it seem as if there is a third set when in fact it is just the main space encompassing everything.

So you really should get:

$\Pr(E \cap O) = \Pr(E) * \Pr(O)\tag{1}$

When they say or hospital claims, it makes me want to write: $\Pr(E \cap O \cap H) = \Pr(E) * \Pr(O) * \Pr(H)$.

Is there something about independence I don't understand? Is it really necessary to state this in the problem?

Once you got (1), I understand you simply plug it in the inclusion exclusion equation and solve for Pr(O):

$\Pr(E \cup O) = \Pr(E) + \Pr(O) - \Pr(E \cap O)\tag{2}$

substitute (1) into (2).

$\Pr(O) = \cfrac{\Pr(E \cup O) - \Pr(E)}{\Pr(E')}= .4$

I appreciate any help. Thank you!

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The "or hospital" is unfortunate wording, and should be disregarded. If you indicate your numerical result I can tell you whether it seems correct. – André Nicolas May 29 '13 at 19:16
Thank you, @AndréNicolas!!!!Is there a reason why you put your answer in the comments instead of the answer space? That way I can close the question as you answered it. Thanks again. – user1527227 May 29 '13 at 19:19

For calculations, there are many ways. I would probably note that $\Pr(O' \cap E')=0.15$. Then from $\Pr(O')\Pr(E')=0.15$ we find that $\Pr(O')=0.6$, and hence $\Pr(O)=0.4$. Your calculation may be a little more efficient.