Probability (theoretical approach)

Toss a coin and you cannot say for certain whether Head or Tail will show — but you can say, with total confidence, that each is equally likely. That confidence is the seed of probability: a number between 0 and 1 that measures how likely an event is, worked out purely by reasoning, without ever performing the experiment. Earlier you met "experimental" probability, built from repeated trials and observed counts. Class 10 introduces the theoretical (classical) approach: count the outcomes that favour an event, count all the equally likely outcomes, and divide. This single idea, applied carefully to coins, dice, cards and marbles drawn from a bag, is the whole of the chapter — along with one powerful shortcut, the complement rule, for when counting "not this" is easier than counting "this."

The core insight

The theoretical probability of an event is the fraction of all equally likely outcomes that favour it, $P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ , always between 0 and 1, with $P(E) + P(\bar E) = 1$ .

The core idea

A random experiment is one whose result cannot be predicted with certainty — tossing a coin, throwing a die, drawing a card. Each possible result is an outcome, and any collection of one or more outcomes is an event. Outcomes are equally likely when nothing favours one over another: every face of a fair die is as likely to land up as any other. The theoretical (classical) probability of an event $E$ is defined by counting outcomes, never by running trials:

P(E) = \frac{\text{number of outcomes favourable to } E}{\text{number of all possible equally likely outcomes}}

where: the numerator counts only the outcomes that satisfy $E$ 's condition, and the denominator counts every outcome the experiment can produce, on the assumption that all of them are equally likely. If outcomes are not equally likely — a biased coin, a loaded die — this formula cannot be applied as taught here. Three facts pin down the range and edge cases of $P(E)$ . Probability is bounded:

0 \le P(E) \le 1

An impossible event (can never happen) has $P(E) = 0$ ; a sure event (must happen) has $P(E) = 1$ . For any event $E$ , its complement $\bar E$ (read "not $E$ ") is the event that $E$ does not occur. Since the favourable outcomes of $E$ and of $\bar E$ together make up every outcome, with no overlap:

P(E) + P(\bar E) = 1

where: $P(\bar E)$ is the probability that $E$ fails to happen. This complement rule is the chapter's most useful shortcut — when "not $E$ " is easier to count than $E$ itself, find $P(\bar E)$ and subtract from 1. Before any formula is applied, the total number of equally likely outcomes must be listed correctly. A single fair coin gives $\{\text{Head}, \text{Tail}\}$ — 2 outcomes. A single fair die gives $\{1,2,3,4,5,6\}$ — 6 outcomes. Two coins tossed together give $\{HH, HT, TH, TT\}$ — 4 outcomes, because $HT$ and $TH$ are genuinely different arrangements even though the coins look alike. Getting this total right, before counting anything favourable, is usually the trickiest step in a probability problem.

Figure

A tree diagram for one die throw showing six equally likely branches from a single starting point, each leading to one face value from 1 to 6

Derivation AWhy the complement rule P(E) + P(not E) equals 1

The goal is to show, from the classical definition of probability alone and without assuming the result, that the probability of an event and the probability of its complement must always add to exactly 1, so that computing one automatically hands you the other by subtraction.

Step 1. Let the experiment have $n$ total equally likely outcomes, and let event $E$ have $f$ favourable outcomes among them, so by definition $P(E) = \dfrac{f}{n}$ .

Step 2. The complement $\bar E$ (" $E$ does not happen") is favoured by every outcome that is not favourable to $E$ . Since the $n$ total outcomes split into exactly two groups — the $f$ that favour $E$ and the rest that favour $\bar E$ — the number of outcomes favouring $\bar E$ is $n - f$ .

Step 3. Apply the classical definition to $\bar E$ using that count: $P(\bar E) = \dfrac{n - f}{n} = \dfrac{n}{n} - \dfrac{f}{n} = 1 - \dfrac{f}{n} = 1 - P(E)$ .

Step 4. Rearranging gives $P(E) + P(\bar E) = 1$ . Nothing here depended on which event $E$ was — the argument used only that "favours $E$ " and "does not favour $E$ " partition the outcomes with no overlap and no gap, so the two probabilities always share the total of 1 between them.

Worked example

Problem. A die is thrown once. Find the probability of getting (a) a prime number, (b) a number greater than 4, (c) a number that is not a multiple of 3.

Step 1. List the sample space and total. The die has 6 equally likely outcomes, $\{1,2,3,4,5,6\}$ , so the total number of outcomes is 6 for every part below.

Step 2. Part (a) — prime number. The primes among 1 to 6 are 2, 3 and 5, giving 3 favourable outcomes: $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$ .

Step 3. Part (b) — greater than 4. Only 5 and 6 qualify, giving 2 favourable outcomes: $P(\text{greater than } 4) = \dfrac{2}{6} = \dfrac{1}{3}$ .

Step 4. Part (c) — not a multiple of 3, using the complement rule. The multiples of 3 among 1 to 6 are 3 and 6, so $P(\text{multiple of }3) = \dfrac{2}{6} = \dfrac{1}{3}$ , and hence $P(\text{not a multiple of }3) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ .

Final answers: $P(\text{prime}) = \dfrac{1}{2}$ , $P(\text{greater than }4) = \dfrac{1}{3}$ , $P(\text{not a multiple of }3) = \dfrac{2}{3}$ .

Competitive stretch

Problem. A box has 3 blue, 2 white and 4 red marbles; one is drawn at random. Find $P(\text{white})$ , $P(\text{not red})$ and $P(\text{blue or white})$ .

Step 1. Total marbles $= 3 + 2 + 4 = 9$ , all equally likely to be drawn, so total outcomes $= 9$ .

Step 2. White: 2 favourable outcomes, so $P(\text{white}) = \dfrac{2}{9}$ . Not red, by the complement rule: $P(\text{red}) = \dfrac{4}{9}$ , so $P(\text{not red}) = 1 - \dfrac{4}{9} = \dfrac{5}{9}$ — matching the direct count of $3 + 2 = 5$ non-red marbles out of 9.

Step 3. Blue or white: favourable outcomes are $3 + 2 = 5$ , so $P(\text{blue or white}) = \dfrac{5}{9}$ . This is the same value as $P(\text{not red})$ , because "not red" and "blue or white" describe exactly the same outcomes here.

Common traps

Miscounting the total outcomes. With two coins or two dice, order matters — $HT$ and $TH$ are different outcomes, not one. Write out the full sample space when unsure.
Treating "favourable" as "special." Favourable outcomes just means outcomes satisfying the stated condition, even a broad one like "not a red marble."
Forgetting equally likely requires it. The classical formula needs equally likely outcomes; it does not apply directly to a biased coin or loaded die at this level.
Skipping the complement rule. Counting "not $X$ " directly is often slower than computing $P(X)$ and subtracting from 1.

Ask the tutor: Ask the AI tutor to walk through listing the sample space, step by step, for a trickier setup like two dice thrown together or a card drawn from a full deck.

Why does the theoretical formula only work when outcomes are equally likely?

The formula counts favourable outcomes out of total outcomes and silently assumes every outcome has the same chance of occurring — otherwise simply counting them tells you nothing about likelihood. A fair die's six faces are equally likely by symmetry, so counting works. A loaded die is not symmetric, so two different faces could have different true chances even though there are still six outcomes; the classical counting formula then gives a wrong answer, and a different method (built from experimental frequencies) would be needed instead.

How do I know whether to count outcomes directly or use the complement rule?

Try both mentally and pick whichever set is easier to list. If the event's own outcomes are few and simple to count — like "prime number" on a die — count them directly. If the event is described as "not something" or covers most of the outcomes — like "not a multiple of 3" or "at least one Head" — it is usually faster to find the probability of the opposite event and subtract from 1, since the opposite event often has far fewer outcomes to count.

Why is HT counted separately from TH when tossing two coins together?

Even though a Head-then-Tail result looks the same as Tail-then-Head when you just glance at "one Head and one Tail," the two coins are distinct physical objects, and swapping which coin shows Head and which shows Tail is a genuinely different arrangement. Listing all four arrangements, $HH$ , $HT$ , $TH$ , $TT$ , as separate equally likely outcomes is what makes each of them have the correct probability of $\frac{1}{4}$ ; merging $HT$ and $TH$ into one outcome would wrongly give it double the chance of $HH$ or $TT$ .

Probability (theoretical approach)

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads