# How do you combine different percents over different instances to find a total percent that something will happen over both instances?

For example a x% that something will happen and a y% that something will happen what is the z% that it will happen(at least once) over both instances

The answer depends on whether your two "somethings" are independent or not. I will give two examples.

Example 1 (Dependence) Let's say that Alice has two chances to draw an ace from a deck of $$10$$ cards that contains $$3$$ aces.

1. If Alice draws a random card once, she will have a $$\frac{3}{10}=0.3=30\%$$ chance to draw an ace and a $$\frac{7}{10}=0.7=70\%$$ chance to not draw an ace.
2. If she does not draw an ace in the first attempt, then $$9$$ cards remain, where $$3$$ of them are aces. So the chance for drawing an ace in the second attempt is $$\frac{3}{9}=\frac{1}{3}\approx0.3333=33.333\%$$ while the chance of failure is $$\frac{6}{9}\approx66.666\%$$.

According to the laws of probability, the chance that Alice does draw an ace is equal to $$100\%$$ minus the chance that she does not. So we have $$P(\text{at least one ace})=1-P(\text{no ace})$$. The probability that Alice draws no ace is given by $$P(\text{no ace})=P(\text{no ace in first attempt})\times P(\text{no ace in second attempt})=0.7\times 0.66666=0.42=42\%$$ so it follows that $$P(\text{at least one ace})=1-0.42=0.58=58\%$$

Example 2 (Independence) Let's say Alice has two chances to roll a $$6$$ on a six-sided die.

1. At the first roll, Alice has a $$\frac{1}{6}\approx0.16667=16.67\%$$ chance to roll a 6 and a $$\frac{5}{6}=0.8333=83.33\%$$ chance not to roll a six.
2. At the second roll, Alice has the same chances since the die does not change.

Here we use the same trick. The chance that Alice does not roll a 6 across both roles is $$P(\text{no six in both attempts})=P(\text{no six in first attempt})\times P(\text{no six in second attempt})=\frac{5}{6}\times \frac{5}{6}=\left(\frac{5}{6}\right)^2=0.6944=69.44\%$$ So the probability that Alice rolls at least one six is $$1-P(\text{no six in both attempts})=1-0.6944=0.3056=30.56\%$$

Summary If there is an $$x$$ chance that event $$e_1$$ will happen and a $$y$$ chance that event $$e_2$$ will happen given that $$e_1$$ didn't happen, the probability that either will happen is given by $$P(e_1\cup e_2)=z=1-(1-x)\times(1-y)=x+y-x\times y$$