Introduction to Probability: Practice

The $4$ equally likely outcomes are HH, HT, TH, TT. 'At least one head' is everything except TT, so $3$ of them.

$P(\text{at least one head}) = 1 - P(\text{TT}) = 1 - \frac{1}{4} = \frac{3}{4}$

Using the complement of 'no heads' is the quick route.

There are $6$ equally likely sectors and $2$ are shaded, so $4$ are not shaded.

$P(\text{not shaded}) = \frac{4}{6} = \frac{2}{3}$

The complement rule agrees: $1 - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}$.

The multiples of $5$ from $1$ to $20$ are $\{5, 10, 15, 20\}$, which is $4$ of the $20$ cards.

$P(\text{multiple of } 5) = \frac{4}{20} = \frac{1}{5}$

Reduce $\frac{4}{20}$ by dividing top and bottom by $4$.

Three coins give $2 \times 2 \times 2 = 8$ equally likely outcomes. The only way to get no heads is all tails (TTT), one outcome.

$P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}$

The complement of 'at least one' is 'none', which is easy to count.

Sums of $11$ or $12$ come from $(5,6), (6,5), (6,6)$, which is $3$ of the $36$ outcomes.

$P(\text{sum} \ge 11) = \frac{3}{36} = \frac{1}{12}$

Reduce $\frac{3}{36}$ to $\frac{1}{12}$.

The bag holds $10$ marbles and $7$ are blue (not red).

$P(\text{not red}) = \frac{7}{10} = 70\%$

The complement rule agrees: $1 - \frac{3}{10} = \frac{7}{10}$.

Each die is even in $3$ ways $\{2, 4, 6\}$, so by the multiplication principle there are $3 \times 3 = 9$ outcomes with both even, out of $36$.

$P(\text{both even}) = \frac{9}{36} = \frac{1}{4}$

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

The numbers greater than $15$ are $\{16, 17, 18, 19, 20\}$, which is $5$ of the $20$ cards.

$P(>15) = \frac{5}{20} = \frac{1}{4}$

Reduce $\frac{5}{20}$ by dividing top and bottom by $5$.

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

$\frac{100}{500} = \frac{1}{5} = 0.2$

One fifth of the trials is $0.2$.

The $4$ equally likely outcomes are HH, HT, TH, TT. Exactly one head means HT or TH, which is $2$ of them.

$P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}$

Reduce $\frac{2}{4}$ to $\frac{1}{2}$.

There are $8$ equally likely sectors and $5$ are shaded.

$P(\text{shaded}) = \frac{5}{8} = 0.625$

Divide $5$ by $8$ to convert the fraction to a decimal.

Count the complement first. Neither die shows a $6$ in $5 \times 5 = 25$ ways, out of $36$.

$P(\text{at least one } 6) = 1 - \frac{25}{36} = \frac{11}{36}$

The complement of 'at least one' is 'none', which is easier to count directly.