probability of any side effects

I know the probability of a range of individual drug side effects. I want to know how to calculate the probability of getting any one of the listed side effects. I'd also like to know how to calculate a given number of selected side effects. To illustrate what I am talking about here's a simple made up example:

A list of three drug side effects provides the probability of getting each one (determined by clinical trials): Probability of Event1 = .01, Probability of Event2 = .15, Probability of Event3 = .02 These are independent of each other (that is, getting one does not affect the probability of getting any other). What are the formulas for calculating:

The probability of getting a side effect (not a specific one, just any one - i.e. the collective probability of getting a side effect);
The probability of getting a specific set of side effects (e.g. 1 and 3).

Thank you very much for your time.

John

jwm649 said:

A list of three drug side effects provides the probability of getting each one (determined by clinical trials): Probability of Event1 = .01, Probability of Event2 = .15, Probability of Event3 = .02 These are independent of each other (that is, getting one does not affect the probability of getting any other). What are the formulas for calculating:
The probability of getting a side effect (not a specific one, just any one - i.e. the collective probability of getting a side effect);
The probability of getting a specific set of side effects (e.g. 1 and 3).

Click to expand...

I really am not sure what to say here.
It seems to me that you need a very basic course in probability.
If \(\displaystyle E_1~\&~E_2\) are truly independent then \(\displaystyle P(E_1\cap E_2)=P(E_1)\cdot P(E_2)\).
The probability that at least one of these occurs is
\(\displaystyle P(E_1\cup E_2\cup E_3)=P(E_1)+P(E_2)+P(E_3)-P(E_1)P(E_2)-\)\(\displaystyle P(E_1)P(E_3)-P(E_3)P(E_2)+P(E_1)P(E_2)P(E_3).\)

Thanks !

Well, years ago I was subjected to probability and statistics but, alas, have forgotten the mechanics of it and no longer have any texts. Thank you very much for your time and assistance.

John