Is my probability correct

Question

Suppose you draw $1$ athlete at random from a group of $100$ athletes such that:

$\{$$30$ swim$\}$, $\{$$44$ run$\}$, $\{$$9$ swim and run$\}$,$\{$$5$ swim, bike and run$\}$, $\{$$11$ swim and bike$\}$, $\{$$10$ bike and run but do not swim$\}$, and $\{$$35$ only bike$\}$.

Let $S, B,$ and $R$ denote the events that the athlete you draw $\text{Swims}$, $\text{Bikes}$, and $\text{Runs}$, respectively. I give the following probabilities: $$P (S \cup R) = \frac{55}{100} + \frac{68}{100} - \frac{9}{100} = \frac{114}{100} $$ $$P (S \cup B) = \frac{55}{100} + \frac{61}{100} - \frac{11}{100} = \frac{105}{100}$$ Did I do something wrong?

Answer 1

No probability can be greater than $1$.

I think, for instance, the $5$ who swim and run are also included among the $30$ who swim, and the $44$ who run.

Answer 2

Probabilities can't be greater than 1. Your approach didn't consider that for example the $5$ who swim and run are also included among the $30$ who swim, and the $44$ who run.

For $P(S \cup R)$: $$ P(S \cup R) = P(S) + P(R) - P(S \cap R) = \frac{30}{100} + \frac{44}{100} - \frac{9}{100} = \frac{65}{100} $$

For $P(S \cup B)$: $$ P(S \cup B) = P(S) + P(B) - P(S \cap B) = \frac{30}{100} + \left( \frac{35 + 10 + 6 + 5}{100} \right) - \left( \frac{6 + 5}{100} \right) = \frac{75}{100} $$ Venn Diagram: