Probability and Sample Spaces

Probability measures how likely something is, on a scale from $0$ (impossible) to $1$ (certain). Before you can work out a probability, you need a clear list of everything that could happen. That list is the sample space, and getting it right is half the battle in every probability problem.

Key ideas

Vocabulary

• An experiment is any process with an uncertain result, like rolling a die.
• An outcome is one possible result, like rolling a $4$.
• The sample space $S$ is the set of all possible outcomes: for one die, $S = \{1, 2, 3, 4, 5, 6\}$.
• An event is a set of outcomes you’re interested in, like “rolling an even number”: $A = \{2, 4, 6\}$.

Discrete and continuous sample spaces

• A discrete sample space has outcomes you can count: the number on a die, the number of goals in a game.
• A continuous sample space has outcomes that are measured and can be any value in a range: a person’s height, the time to run $100$ m.

Theoretical probability

When all outcomes are equally likely:

$$P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}$$

Every probability is between $0$ and $1$: $0 \le P(A) \le 1$. Probabilities can be written as fractions, decimals, or percentages.

Listing a sample space

• List the outcomes for simple experiments.
• Use a table for two-step experiments, like rolling two dice.
• Use a tree diagram for several steps, like flipping three coins.

The sums of two dice, for example:

$+$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

There are $6 \times 6 = 36$ equally likely outcomes.

Probability distributions

A probability distribution lists every outcome with its probability. Because one of the outcomes must happen, the probabilities always add up to $1$.

Worked examples

Example 1: One die

A fair die is rolled. Find $P(\text{even})$ and $P(\text{greater than } 4)$.

Solution. $S = \{1, 2, 3, 4, 5, 6\}$, so $n(S) = 6$.

$$P(\text{even}) = \frac{3}{6} = \frac{1}{2}, \qquad P(\text{greater than } 4) = \frac{2}{6} = \frac{1}{3}$$

Example 2: Two dice

Two fair dice are rolled. Use the table above to find $P(\text{sum is } 7)$ and $P(\text{sum is } 10 \text{ or more})$.

Solution. A sum of $7$ appears $6$ times in the table (along a diagonal):

$$P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}$$

Sums of $10$, $11$, and $12$ appear $3 + 2 + 1 = 6$ times:

$$P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$$

Example 3: A probability distribution

Three fair coins are flipped. Make a probability distribution for the number of heads.

Solution. A tree diagram gives $2 \times 2 \times 2 = 8$ equally likely outcomes:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Number of heads	$0$	$1$	$2$	$3$
Probability	$\tfrac{1}{8}$	$\tfrac{3}{8}$	$\tfrac{3}{8}$	$\tfrac{1}{8}$

Check: $\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{3}{8} + \tfrac{1}{8} = 1$. ✓

Example 4: A missing probability

A spinner’s outcomes have probabilities $0.2$, $0.35$, $0.3$, and $x$. Find $x$.

Solution. The probabilities must add to $1$:

$$0.2 + 0.35 + 0.3 + x = 1 \quad\Rightarrow\quad x = 0.15$$

Common mistakes

Assuming outcomes are equally likely when they aren’t. The sums of two dice range from $2$ to $12$, but they aren’t equally likely: a sum of $7$ is six times as likely as a sum of $2$. List the $36$ equally likely pairs instead.

Counting $(2, 5)$ and $(5, 2)$ as the same outcome. With two dice, they’re different outcomes (think of one die as red and one as blue).

A probability greater than $1$, or negative. That always means a mistake.

A distribution that doesn’t add to $1$. Check the total every time.

Practice

1. (Warm-up) One card is drawn from a standard $52$-card deck. Find $P(\text{heart})$, $P(\text{face card})$, and $P(\text{red ace})$.

Solution

$P(\text{heart}) = \tfrac{13}{52} = \tfrac{1}{4}$. $P(\text{face card}) = \tfrac{12}{52} = \tfrac{3}{13}$ (jack, queen, king in four suits). $P(\text{red ace}) = \tfrac{2}{52} = \tfrac{1}{26}$.

2. (Warm-up) A spinner has $8$ equal sections numbered $1$ to $8$. Find $P(\text{prime number})$.

Solution

The primes are $2, 3, 5, 7$: $P = \tfrac{4}{8} = \tfrac{1}{2}$.

3. (Warm-up) Is each sample space discrete or continuous?

• (a) the number of goals in a hockey game
• (b) the mass of an apple
• (c) the number of students absent today
• (d) the temperature at noon

Solution

(a) Discrete. (b) Continuous. (c) Discrete. (d) Continuous.

4. (Core) Two fair dice are rolled. Find $P(\text{doubles})$ and $P(\text{sum is } 5)$.

Solution

Doubles: $(1,1), \dots, (6,6)$, so $\tfrac{6}{36} = \tfrac{1}{6}$.

Sum $5$: $(1,4), (2,3), (3,2), (4,1)$, so $\tfrac{4}{36} = \tfrac{1}{9}$.

5. (Core) A probability distribution has probabilities $0.15$, $0.4$, $x$, and $0.25$. Find $x$.

Solution

$0.15 + 0.4 + x + 0.25 = 1$, so $x = 0.2$.

6. (Core) A family has three children. Assuming each child is equally likely to be a boy or a girl, find $P(\text{exactly two girls})$.

Solution

There are $8$ equally likely outcomes (like the coins in Example 3), and $3$ of them have exactly two girls (GGB, GBG, BGG): $P = \tfrac{3}{8}$.

7. (Core) A bag holds $5$ red, $3$ blue, and $2$ green marbles. One is drawn. Find $P(\text{not red})$ and $P(\text{blue or green})$.

Solution

Both events are the same five marbles: $P = \tfrac{5}{10} = \tfrac{1}{2}$.

8. (Challenge) Two fair dice are rolled, and $X$ is the larger of the two numbers (or the shared number for doubles). Find the probability distribution of $X$.

Solution

For $X = k$, at least one die shows $k$ and neither is bigger. Counting in the table: $X = 1$ has $1$ outcome, $X = 2$ has $3$, $X = 3$ has $5$, and in general $X = k$ has $2k - 1$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P$	$\tfrac{1}{36}$	$\tfrac{3}{36}$	$\tfrac{5}{36}$	$\tfrac{7}{36}$	$\tfrac{9}{36}$	$\tfrac{11}{36}$

Check: $1 + 3 + 5 + 7 + 9 + 11 = 36$. ✓

9. (Challenge) Two fair dice are rolled and the numbers are multiplied. Find $P(\text{product is even})$.

Solution

The product is odd only if both numbers are odd: $3 \times 3 = 9$ outcomes. So $27$ of the $36$ outcomes give an even product:

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