Imagine we have a pie and let $p,q$ be ratios between 0 and 1 (non-inclusive). If I first take $p$ out of the pie, and then $q$ out of the remainder, how much is that of the whole pie?

I can easily visualize that if $p=0.5$ so I take half of the pie, and then take again $q=0.5$ of the remainder, I will have taken a total of 0.75 of the pie. But how do I compute this in general for any $p$ and $q$?

  • $\begingroup$ After you take out $p$, how much is left? After you take $q$ of what's left, how much have you taken in total? $\endgroup$ – NickD Jul 26 '18 at 19:46

The rule you want is $$ pq $$ when you take fraction $p$ and then fraction $q$ of what you just took.

In your problem you take $p$ of the pie and then $q$ from what's left over so your total is $$ p + (1-p)q . $$


An alternate way to arrive at the same answer is to calculate what's left after each step: After taking p out of the pie, you have (1-p) of the pie left. You then take q out of that, so you'd have (1-p)(1-q) of the pie left. In your example case, you get (1-.5)(1-.5) =.5^2 = .25 left.

To calculate how much you'd taken, you'd subtract this from one, so you'd have 1-(1-p)(1-q) that you took. If you simplify this, you get the same answer that Ethan gave.

This method is generally useful to learn because it has applications in a lot of situations, especially probability.


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