Sample space and events

Every probability problem starts with an experiment , a process that produces an outcome that cannot be predicted with certainty. Tossing a coin, rolling a die, drawing a card, drawing a ball from a bag , these are all probability experiments. To compute probabilities, we begin by listing all the possible outcomes.

Sample space

The sample space of an experiment is the set of all possible outcomes. It is denoted $S$.

Examples:

• Tossing one coin: $S = \{H, T\}$. Size $2$.
• Tossing two coins: $S = \{HH, HT, TH, TT\}$. Size $4$.
• Rolling one die: $S = \{1, 2, 3, 4, 5, 6\}$. Size $6$.
• Rolling two dice: $S = \{(1,1), (1,2), \ldots, (6,6)\}$. Size $36$.
• Drawing one card from a deck: $S =$ all $52$ cards. Size $52$.
• Drawing one ball from a bag with $5$ red and $3$ blue balls: $S =$ each of the $8$ balls (we treat each ball as distinct even if they look alike). Size $8$.

Equally likely outcomes

Two outcomes are equally likely if there is no reason to expect one more than the other. For a fair coin, $H$ and $T$ are equally likely. For a fair die, each of $1, 2, 3, 4, 5, 6$ is equally likely.

When the outcomes are equally likely, the probability of any single outcome is $1/(\text{total outcomes})$.

For experiments with non-equally-likely outcomes, the formula gets more nuanced (covered in higher classes). For Class X, almost every problem assumes equally likely outcomes.

Events

An event is any subset of the sample space , i.e., any collection of outcomes. We typically describe an event in words ("getting an even number") and then translate to an explicit list.

For "rolling a die, getting an even number": event $E = \{2, 4, 6\}$. Size $3$.

For "rolling two dice, getting a sum of $7$": event $E = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$. Size $6$.

Probability of an event

When all outcomes are equally likely,

$P(E) = \frac{\text{number of outcomes in $E$}}{\text{total number of outcomes in $S$}}.$

For "rolling a die, getting an even number": $P(E) = 3/6 = 1/2$.

For "rolling two dice, sum $7$": $P(E) = 6/36 = 1/6$.

For "drawing a king from a deck": there are $4$ kings out of $52$ cards. $P = 4/52 = 1/13$.

Sure and impossible events

• A sure event is one that always happens. $P = 1$. E.g., "drawing any card from a deck".
• An impossible event is one that never happens. $P = 0$. E.g., "drawing a king of red apples from a deck".

For any event $E$, $0 \leq P(E) \leq 1$. Probabilities outside this range are nonsensical.

Worked examples

Example 1. A die is rolled. Find the probability of getting (a) an even number, (b) a multiple of $3$, (c) a prime number.

(a) Favourable: $\{2, 4, 6\}$. $P = 3/6 = 1/2$.

(b) Favourable: $\{3, 6\}$. $P = 2/6 = 1/3$.

(c) Primes are $2, 3, 5$. $P = 3/6 = 1/2$.

Example 2. Two coins are tossed simultaneously. Find the probability of getting (a) two heads, (b) at least one head, (c) no head.

Sample space: $\{HH, HT, TH, TT\}$.

(a) $P(\text{two heads}) = 1/4$.

(b) $P(\geq 1 H) = 3/4$ ($HH, HT, TH$).

(c) $P(\text{no head}) = 1/4$ ($TT$).

Example 3. A bag contains $5$ red, $4$ green, and $3$ blue balls. One ball is drawn. Find the probability of getting (a) red, (b) green, (c) not blue.

Total: $12$ balls.

(a) $P(\text{red}) = 5/12$.

(b) $P(\text{green}) = 4/12 = 1/3$.

(c) $P(\text{not blue}) = (5 + 4)/12 = 9/12 = 3/4$.

Example 4. A card is drawn from a well-shuffled deck of $52$. Find the probability that it is (a) a heart, (b) a king, (c) the king of hearts, (d) a red card.

(a) Hearts: $13$. $P = 13/52 = 1/4$.

(b) Kings: $4$. $P = 4/52 = 1/13$.

(c) King of hearts: $1$. $P = 1/52$.

(d) Red cards: $26$ (hearts + diamonds). $P = 26/52 = 1/2$.

Example 5. Two dice are rolled. Find the probability that the sum is (a) $9$, (b) at most $4$, (c) a multiple of $5$.

Sample space size: $36$.

(a) Sum $9$: $(3,6), (4,5), (5,4), (6,3) \Rightarrow 4$. $P = 4/36 = 1/9$.

(b) Sum $\leq 4$: $(1,1), (1,2), (2,1), (1,3), (3,1), (2,2) \Rightarrow 6$. $P = 6/36 = 1/6$.

(c) Multiples of $5$ (i.e., $5$ or $10$): for $5$, $(1,4),(2,3),(3,2),(4,1) \Rightarrow 4$. For $10$, $(4,6),(5,5),(6,4) \Rightarrow 3$. Total $7$. $P = 7/36$.

Try it yourself

1. A die is rolled. Find the probability of getting (a) a $4$, (b) a number greater than $4$, (c) an odd prime.
2. Two coins are tossed. Find $P($exactly one head$)$.
3. Three coins are tossed. Sample space has size $8$. Find $P($two heads$)$.
4. A bag has $7$ white and $5$ black balls. $P($white$) = ?$
5. A card is drawn. Find $P($ace$)$, $P($spade$)$, $P($face card$)$.
6. Two dice. Find $P($sum is $11$ or 12$$).
7. Two dice. Find $P($doubles, i.e., both dice show the same number$)$.
8. A box has $10$ cards numbered $1$ to $10$. One is drawn. $P($prime number$) = ?$
9. A bag has tickets numbered $1$ to $50$. $P($multiple of 7$$) = ?
10. A two-digit number is randomly chosen from $10, 11, \ldots, 99$. $P($number divisible by 5$$) = ?
11. A letter is chosen from "MATHEMATICS". $P($vowel$) = ?$
12. A die is rolled. $P($not 6$$) = ?

Pitfalls / Insight

(1) "Two coins are tossed simultaneously" produces $4$ outcomes ($HH, HT, TH, TT$), not $3$. The order matters when listing the sample space, even if the two coins look identical.

(2) For two dice, similarly, $(1, 2)$ and $(2, 1)$ are different outcomes. Sample space has $36$ ordered pairs.

(3) For "at least one X" probability, it is often easier to compute $P($no X$)$ and subtract from $1$.

(4) Probability is a ratio, not a count. Express your answer as a fraction in lowest terms (e.g., $1/2$, not $26/52$).