Introduction to Probability: Additional Practice (Set 1)

Of the $36$ equally likely outcomes, the pairs adding to $5$ are $(1,4), (2,3), (3,2), (4,1)$, which is $4$.

$P(\text{sum } 5) = \frac{4}{36} = \frac{1}{9}$

Reduce $\frac{4}{36}$ to $\frac{1}{9}$.

The multiples of $3$ on a die are $\{3, 6\}$, which is $2$ of the $6$ faces.

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

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

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

$P(\text{not win}) = 1 - \frac{2}{5} = \frac{3}{5}$

The two probabilities add to $1$.

Divide the number of times it happened by the number of trials.

$\frac{80}{400} = \frac{1}{5}$

Divide top and bottom by $80$ to reach $\frac{1}{5}$.

There are $15$ equally likely marbles and $9$ are red.

$P(\text{red}) = \frac{9}{15} = \frac{3}{5}$

Reduce $\frac{9}{15}$ by dividing top and bottom by $3$.

Equal sectors are equally likely, so there are $12$ equally likely outcomes, $3$ of them shaded.

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

Reduce $\frac{3}{12}$ by dividing top and bottom by $3$.

The product is odd only when both dice are odd, which happens in $3 \times 3 = 9$ ways. So the product is even in $36 - 9 = 27$ ways.

$P(\text{product even}) = \frac{27}{36} = \frac{3}{4}$

Reduce $\frac{27}{36}$ by dividing top and bottom by $9$.

There are $5$ multiples of $4$, so $20 - 5 = 15$ numbers are not multiples of $4$.

$P(\text{not a multiple of } 4) = \frac{15}{20} = \frac{3}{4}$

The complement rule agrees: $1 - \frac{5}{20} = \frac{15}{20} = \frac{3}{4}$.

There are $8$ equally likely outcomes. Exactly two heads happens as HHT, HTH, and THH, which is $3$ of them.

$P(\text{exactly two heads}) = \frac{3}{8}$

List the eight outcomes and count the ones with two heads.

A die only shows the numbers $1$ through $6$, so a $7$ can never come up. It is an impossible event.

$P(7) = \frac{0}{6} = 0$

An impossible event has probability $0$.

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

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

Reduce $\frac{4}{12}$ to $\frac{1}{3}$.

Divide the number of times it happened by the number of trials, then convert.

$\frac{250}{1000} = \frac{1}{4} = 25\%$

One quarter of the trials is $25\%$.