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Independent events Example: if the chance of having blond hair is 0.3 and the chance of having a cold is 0.2, the chance of meeting a blond-haired person with a cold is: 0.3x0.2=0.06

Not mutually exclusive events Example: if the chance of having diabetes is 10% and the chance of being obese is 30%, the chance of meeting someone who is obese or has diabetes or both is (0.1+0.3)-0.1x0.3=0.37

I have question: What is the difference between examples above? Why in one case we use the formula of Independent events but in another formula of not mutually exclusive events? I found these examples is similar. I have only one guess: obesity is risk factor of diabetes mellitus and fat tissue has influence on glucose metabolism. So, obesity can tell us something about the occurrence of diabetes. Is it right guess? And if it right... Why we should substract out? I think, probability to find person with obesity+diabetes is higher than if it would Independent conditions.

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  • $\begingroup$ In the first example, you have $\Pr(A \cap B)$, and in the second example, you are computing $\Pr(A \cup B)$. In both cases, you used the independence of $A$ and $B$; i.e., $\Pr(A \cap B)=\Pr(A)\cdot \Pr(B)$. $\endgroup$ – Math Lover Oct 17 '17 at 16:18
  • $\begingroup$ Just to be clear, to make the arguments you are making require assumptions you never stated. For both you are assuming independence. That's not an obvious assumption. Indeed it is seriously false for the second example (there is a positive correlation between obesity and diabetes). For a math problem you can assume whatever you want, but you should state such assumptions clearly, especially when they contradict reality. $\endgroup$ – lulu Oct 17 '17 at 16:24
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In your second example you use the formula:

$$P(D \cup O) = P(D) + P(O) - P(D \cap O)$$

and you also used:

$$P(D \cap O) = P(D) \cdot P(O)$$

The first formula is always true, but the second assumes that you are dealing with two independent events, which you can understand as follows:

It is always true that:

$$P(A \cap B) = P(A) \cdot P(B|A)$$

But if $A$ and $B$ are independent, you have that:

$$P(B|A) = P(B)$$

Plugging that in, we thus get:

$$P(A \cap B) = P(A) \cdot P(B|A) = P(A) \cdot P(B)$$

So yes, both examples assumed that the events are independent. ... though the independence of the second example is probably not true in real life ...

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The reason for the difference is that in one case you want blond AND cold; but in the other case you want diabetes OR obese.

If you did blond OR cold, the computation would look similar to the diabetes OR obese calculation. [But see below.]

However, you are treating diabetes and obesity as independent characteristics, which is presumably incorrect.

Blond and cold are likely but not surely independent characteristics.

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