Probability of single events

Flip a coin, roll a die, draw a card, pick a ball from a bag — every one of these is a random experiment: the result cannot be predicted in advance, though every possible result is known ahead of time. Probability is the branch of arithmetic that turns "how likely" into an exact number between $0$ and $1$ , by comparing the outcomes we want against every outcome that could happen. ICSE Class 10 keeps this tightly scoped: one experiment at a time (a single coin, a single die, one card, one ball), using the classical definition of probability, which counts equally likely outcomes rather than relying on repeated trials. Two coins tossed together, conditional probability, and the formal addition/multiplication theorems belong to ISC (Classes 11–12) and are out of syllabus here. This article expands the core definition, walks through fully worked die and bag problems, flags the traps examiners love to set, and gives quiz and exam-card practice — all built on the classical formula $P(E) = n(E)/n(S)$ .

The core insight

For a single random experiment with equally likely outcomes, the probability of an event is the count of favourable outcomes divided by the total count of possible outcomes, $P(E) = n(E)/n(S)$ , and this always lies between $0$ (impossible) and $1$ (certain).

The core idea

A random experiment has a known set of possible results but an unpredictable actual result — tossing a coin, rolling a die, drawing a card. The sample space $S$ is the set of all possible outcomes. Tossing one coin gives $S = \{H, T\}$ , so $n(S) = 2$ . Rolling one ordinary die gives $S = \{1,2,3,4,5,6\}$ , so $n(S) = 6$ . Drawing one card from a well-shuffled standard deck gives $n(S) = 52$ . An event $E$ is the particular subset of outcomes we care about. "Getting an even number" on one die roll is $E = \{2,4,6\}$ , so $n(E) = 3$ . Every event is judged against the same fixed sample space of its experiment. Provided every outcome in $S$ is equally likely — no outcome is favoured over another — the classical (theoretical) probability of $E$ is:

P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{number of outcomes favourable to } E}{\text{total number of possible outcomes}}

where: $n(E)$ is the count of outcomes in the event, and $n(S)$ is the count of outcomes in the whole sample space. Three facts follow immediately and are used constantly to sanity-check answers:

$0 \le P(E) \le 1$ always — a probability outside this range signals a counting mistake.
$P(E) = 0$ for an impossible event (it can never happen); $P(E) = 1$ for a certain event (it must happen).
If $E'$ is the complementary event (" $E$ does not happen"), then

P(E) + P(E') = 1, \qquad P(E') = 1 - P(E)

where: $E'$ (also written $\bar{E}$ ) contains every outcome of $S$ that is not in $E$ . This complement rule is the fastest route through "not", "at least one", and "other than" style questions, because it is often easier to count what you don't want than what you do. The working method is always the same order: write $S$ out fully so $n(S)$ is correct, identify exactly which outcomes of $S$ satisfy the event so $n(E)$ is correct, then divide and simplify the fraction.

Figure

A probability number line from zero at impossible through one half at even chance to one at certain, with a die and a coin marked at their positions

Derivation AWhy P(E) = n(E)/n(S) is the right count for a single equally likely event

Show that if a random experiment has $n(S)$ total outcomes and every outcome is equally likely to occur, then the fraction of trials in which an event $E$ occurs, in the long run, matches the ratio $n(E)/n(S)$ , so this ratio is the correct definition of the event's probability.

Step 1. Since every outcome in the sample space $S$ is equally likely, each individual outcome carries the same "share" of the total certainty. There are $n(S)$ outcomes in total and they must together account for all of the probability, since the experiment must produce some outcome.

Step 2. Total probability across the whole sample space is $1$ (certainty), because the experiment is certain to land on one of its listed outcomes. Splitting this $1$ equally among the $n(S)$ equally likely outcomes gives each single outcome a probability of $\dfrac{1}{n(S)}$ .

Step 3. An event $E$ is simply a collection of $n(E)$ of these individual outcomes — the ones that satisfy whatever condition defines $E$ . Since probabilities of distinct outcomes add up when we ask for "any one of these outcomes", the probability of $E$ is the sum of $n(E)$ equal pieces, each worth $\dfrac{1}{n(S)}$ .

Step 4. Adding $n(E)$ copies of $\dfrac{1}{n(S)}$ gives $P(E) = n(E) \times \dfrac{1}{n(S)} = \dfrac{n(E)}{n(S)}$ , which is exactly the classical formula. This confirms why the "equally likely" condition is essential — the argument only works because every outcome contributes the same $\dfrac{1}{n(S)}$ ; an unevenly weighted experiment would need different reasoning entirely.

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. Write the sample space. Throwing one die gives $S = \{1,2,3,4,5,6\}$ , so $n(S) = 6$ .

Step 2. Part (a). The prime numbers inside $S$ are $2, 3, 5$ , so $E = \{2,3,5\}$ and $n(E) = 3$ . Then $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$ .

Step 3. Part (b). The outcomes greater than $4$ inside $S$ are $5, 6$ , so $n(E) = 2$ . Then $P(\text{greater than } 4) = \dfrac{2}{6} = \dfrac{1}{3}$ .

Step 4. Part (c). Multiples of $3$ inside $S$ are $3, 6$ , so $n(E) = 2$ and $P(\text{multiple of }3) = \dfrac{2}{6} = \dfrac{1}{3}$ . By the complement rule, $P(\text{not a multiple of }3) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ . Direct check: numbers that are not multiples of $3$ are $1,2,4,5$ — four outcomes, giving $P = \dfrac{4}{6} = \dfrac{2}{3}$ , which matches.

Final answers: P(prime) = 1/2, P(greater than 4) = 1/3, P(not a multiple of 3) = 2/3.

Competitive stretch

Problem. A bag has 5 red, 4 green, and 3 blue balls, identical apart from colour. One ball is drawn at random. Find P(red or green).

Step 1. Total balls: $n(S) = 5+4+3 = 12$ , and since the balls are identical apart from colour, the classical definition applies.

Step 2. Favourable outcomes for "red or green" are $5 + 4 = 9$ , so $P(\text{red or green}) = \dfrac{9}{12} = \dfrac{3}{4}$ . This also equals $1 - P(\text{blue}) = 1 - \dfrac{3}{12} = \dfrac{3}{4}$ by the complement rule, since "red or green" is exactly "not blue" here — confirming the final answer: 3/4.

Common traps

Miscounting the sample space. Forgetting an outcome — leaving out 1 among primes, or treating a die as having five faces instead of six — throws off $n(S)$ before the question even starts. Always list $S$ explicitly.
Confusing an event with its complement. For "not"/"at least one" questions, students often compute $P(E)$ correctly but forget to subtract from $1$ , or subtract the wrong number entirely.
Assuming unequal outcomes are equally likely. $P(E) = n(E)/n(S)$ only holds when every individual outcome has the same chance; an unevenly sized spinner sector needs area or angle, not a simple outcome count.
Leaving the fraction unsimplified, or reaching for two-event reasoning. Simplify the final fraction, and remember this syllabus covers only single events — two coins or two dice tossed together belongs to compound-event probability, which is out of scope here.

Ask the tutor: Want more practice turning a word problem into $n(E)$ and $n(S)$ , or a check on where you're going wrong with a complement-rule question? Ask the AI tutor for extra single-event probability problems with step-by-step solutions.

Why does P(E) plus P(E') always equal 1, and not something else

Because $E$ and $E'$ together account for every single outcome in the sample space $S$ , with no overlap and nothing left out — every outcome either satisfies $E$ or it does not, there is no third option. Since the whole sample space carries total probability $1$ (something is certain to happen), splitting it into these two non-overlapping pieces means their probabilities must add to exactly $1$ .

How do I know when to use the complement rule instead of counting E directly

Reach for the complement rule whenever the event is phrased with "not", "other than", or "at least one out of a short list" — these are often easier to count as the leftover than the target itself. For example, "not blue" from a bag of many colours is quicker as $1 - P(\text{blue})$ than listing every non-blue colour and adding them up. If direct counting is just as easy (as with "prime number" on a die), there is no need to detour through the complement.

Can probability ever come out negative or bigger than 1, and what does that mean

No — for a valid single event, $0 \le P(E) \le 1$ always, because $n(E)$ can never be negative and can never exceed $n(S)$ (an event's outcomes are a subset of the sample space, never more). If your calculation gives a negative number or something above $1$ , it is a sure sign of a counting mistake — either $n(S)$ was undercounted, or $n(E)$ was overcounted by including an outcome that is not actually in $S$ .

Probability of single events

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads