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


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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 $$P(e_1\cup e_2)=z=1-(1-x)\times(1-y)=x+y-x\times y$$


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