Introduction to Probability: Practice

Equal sectors are equally likely, so there are $8$ equally likely outcomes, $3$ of them shaded.

$P(\text{shaded}) = \frac{3}{8}$

The fraction $\frac{3}{8}$ is already in simplest form.

The bag holds $4 + 8 = 12$ marbles, all equally likely, and $4$ are red.

$P(\text{red}) = \frac{4}{12} = \frac{1}{3}$

Reduce $\frac{4}{12}$ by dividing top and bottom by $4$.

Rain and no rain are complements, so their probabilities add to $1$.

$P(\text{no rain}) = 1 - 0.3 = 0.7$

The complement rule works for any event, not only equally likely ones.

Rewrite the fraction with denominator $100$ to read off the percent.

$\frac{3}{5} = \frac{60}{100} = 60\%$

Multiply top and bottom by $20$ to get $\frac{60}{100}$.

The numbers that are not greater than $4$ are $\{1, 2, 3, 4\}$, which is $4$ favorable outcomes out of $6$.

$P(\le 4) = \frac{4}{6} = \frac{2}{3}$

Reduce $\frac{4}{6}$ to $\frac{2}{3}$.

Experimental probability is the number of times it happened over the number of trials.

$P_{\text{exp}}(\text{red}) = \frac{20}{50} = \frac{2}{5}$

Reduce $\frac{20}{50}$ by dividing top and bottom by $10$.

Winning and not winning are complements, so use the complement rule.

$P(\text{not win}) = 1 - \frac{1}{4} = \frac{3}{4}$

The two probabilities add to $1$.

The multiples of $3$ from $1$ to $10$ are $\{3, 6, 9\}$, which is $3$ favorable outcomes out of $10$.

$P(\text{multiple of } 3) = \frac{3}{10}$

The fraction $\frac{3}{10}$ is already in simplest form.

Two of the six faces, $\{1, 2\}$, are favorable.

$P(1 \text{ or } 2) = \frac{2}{6} = \frac{1}{3}$

Reduce $\frac{2}{6}$ to $\frac{1}{3}$.

Equal sectors are equally likely, so there are $10$ equally likely outcomes, $4$ of them blue.

$P(\text{blue}) = \frac{4}{10} = \frac{2}{5}$

Reduce $\frac{4}{10}$ by dividing top and bottom by $2$.

The bag holds $3 + 5 + 4 = 12$ marbles. Not green means red or blue, which is $3 + 5 = 8$.

$P(\text{not green}) = \frac{8}{12} = \frac{2}{3}$

The complement rule agrees: $1 - \frac{4}{12} = \frac{8}{12} = \frac{2}{3}$.

The numbers greater than $2$ are $\{3, 4, 5, 6\}$, which is $4$ favorable outcomes out of $6$.

$P(>2) = \frac{4}{6} = \frac{2}{3}$

Reduce $\frac{4}{6}$ to $\frac{2}{3}$.