Probability

"It will probably rain" is a hunch; $P(\text{rain}) = 0.7$ is a claim you can check. Probability is the branch of mathematics that turns the vague language of chance into numbers between 0 and 1 that can be calculated, compared, and combined. In ISC Class 11, that means building the vocabulary of random experiments, sample spaces and events, then pinning down exactly what a probability number means through the classical definition and the addition theorem for "or" events. This is deliberately elementary, classical probability: no conditioning on new information, no independence, no distributions — those arrive in Class XII. What you build here is the toolkit every later chapter on probability rests on: how to count a sample space correctly, and how to combine the probabilities of two events without double-counting the outcomes they share.

The core insight

A probability is the fraction of equally likely outcomes that favour an event, and the probability of " $A$ or $B$ " is $P(A)+P(B)$ minus whatever outcomes $A$ and $B$ share, so nothing gets counted twice.

The core idea

A random experiment is one whose outcome cannot be predicted in advance, though every possible outcome is known beforehand — tossing a coin or rolling a die, not "will it rain in July." The sample space $S$ is the set of all possible outcomes, and an event $E$ is any subset of $S$ : it occurs exactly when the actual outcome lands inside that subset. A short vocabulary carries the whole chapter. The sure event is $S$ itself; the impossible event is $\varnothing$ . Two events are mutually exclusive if they cannot both occur, $A \cap B = \varnothing$ . A family of events is exhaustive if their union covers all of $S$ . "Not $A$ " is the complement $A'$ ; " $A$ and $B$ " is $A \cap B$ ; " $A$ or $B$ " is $A \cup B$ .

Axiomatic definition. Probability assigns every event $E \subseteq S$ a number $P(E)$ with

0 \le P(E) \le 1, \qquad P(S) = 1,

where: $P(E)$ is the probability of $E$ , and mutually exclusive events add: if $A \cap B = \varnothing$ then $P(A \cup B) = P(A) + P(B)$ .

Classical definition. When every outcome in a finite sample space is equally likely,

P(E) = \frac{n(E)}{n(S)},

where: $n(E)$ is the number of outcomes favourable to $E$ and $n(S)$ is the total number of outcomes. This single formula is the workhorse of the whole chapter — every dice, coin and card problem reduces to counting $n(E)$ and $n(S)$ correctly. Two identities follow directly. First, since $E$ and its complement $E'$ are always mutually exclusive and always exhaustive,

P(E') = 1 - P(E).

Second, for any two events $A$ and $B$ , not necessarily mutually exclusive, the addition theorem states

P(A \cup B) = P(A) + P(B) - P(A \cap B).

Figure

Two overlapping circles labelled A and B inside a rectangle labelled S, with the overlapping region shaded to show it is counted once not twice

When $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$ and the addition theorem collapses to plain addition, $P(A \cup B) = P(A) + P(B)$ — the special case hiding inside the general rule.

Derivation AWhere the addition theorem comes from

The goal is to show, using nothing but the classical definition and simple counting, why the formula for " $A$ or $B$ " must subtract $P(A \cap B)$ rather than just adding $P(A)$ and $P(B)$ , so that outcomes lying in both events are never counted twice when we compute the probability of their union.

Step 1. Start from counting, not probability. The event $A \cup B$ occurs on every outcome that lies in $A$ , in $B$ , or in both. If you simply add $n(A) + n(B)$ , every outcome that lies in both $A$ and $B$ gets counted twice — once inside $n(A)$ and once inside $n(B)$ — while outcomes in only one of the events are counted correctly, once.

Step 2. Correct the double count. The set of outcomes counted twice is exactly $A \cap B$ , so subtracting $n(A \cap B)$ once removes exactly one of the two extra counts, leaving every outcome counted exactly once: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ .

Step 3. Convert counts to probabilities. Divide the whole equation by $n(S)$ , the total number of equally likely outcomes, and each term becomes the classical probability of that event: $\dfrac{n(A \cup B)}{n(S)} = \dfrac{n(A)}{n(S)} + \dfrac{n(B)}{n(S)} - \dfrac{n(A \cap B)}{n(S)}$ , which is exactly $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Step 4. Check the special case. If $A$ and $B$ are mutually exclusive, $A \cap B = \varnothing$ , so $n(A \cap B) = 0$ and the formula reduces to $P(A \cup B) = P(A) + P(B)$ , matching the addition rule stated in the axioms. The general theorem is therefore not a new assumption — it is forced by correct counting whenever events overlap.

Worked example

A card is drawn at random from a well-shuffled standard deck of 52 playing cards. Find the probability that the card drawn is (a) not a king, and (b) a heart or a queen.

Step 1. Set up the sample space: $n(S) = 52$ , every card equally likely.

Step 2. Event $E$ = "card is a king." There are 4 kings, so $n(E) = 4$ and $P(E) = \frac{4}{52} = \frac{1}{13}$ . The complement rule gives $P(E') = 1 - \frac{1}{13} = \frac{12}{13}$ for "not a king."

Step 3. Let $A$ = "card is a heart" and $B$ = "card is a queen." There are 13 hearts, so $P(A) = \frac{13}{52} = \frac{1}{4}$ ; there are 4 queens, so $P(B) = \frac{4}{52} = \frac{1}{13}$ . $A$ and $B$ overlap only at the queen of hearts, so $n(A \cap B) = 1$ and $P(A \cap B) = \frac{1}{52}$ .

Step 4. Apply the addition theorem: $P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$ . The probability of "not a king" is $\frac{12}{13}$ , and the probability of "a heart or a queen" is $\frac{4}{13}$ .

Competitive stretch

A fair die is rolled once. Let $A$ = "number is even" and $B$ = "number is greater than 4." Since $S = \{1,2,3,4,5,6\}$ , $A = \{2,4,6\}$ gives $P(A) = \frac{1}{2}$ , and $B = \{5,6\}$ gives $P(B) = \frac{1}{3}$ . Their overlap is $A \cap B = \{6\}$ , so $P(A \cap B) = \frac{1}{6}$ , and the addition theorem gives $P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}$ — matching a direct count of $A \cup B = \{2,4,5,6\}$ , which is $\frac{4}{6} = \frac{2}{3}$ .

Common traps

Adding $P(A)$ and $P(B)$ without subtracting $P(A \cap B)$ whenever the events are not mutually exclusive — this double-counts the shared outcomes, as with the queen of hearts above.
Confusing mutually exclusive with exhaustive. Mutually exclusive means $A \cap B = \varnothing$ (cannot happen together); exhaustive means $A \cup B = S$ (together cover everything). An event pair can be one, both, or neither.
Miscounting $n(S)$ for compound experiments such as two dice or drawing without replacement — always write out or clearly justify the sample space before counting favourable outcomes.
Treating $P(E') = 1 - P(E)$ as a special-case shortcut. It holds for every event $E$ , because $E$ and $E'$ are always mutually exclusive and always exhaustive by the very definition of a complement.

Ask the tutor: if you're unsure whether two events in a problem are mutually exclusive, or want more practice setting up sample spaces for dice, coin, or card problems, ask the AI tutor to work through it with you.

Why do we subtract P of A and B instead of just adding P of A and P of B

Because any outcome that lies in both $A$ and $B$ gets counted once inside $P(A)$ and once again inside $P(B)$ , so a plain sum overstates the true probability of " $A$ or $B$ ." Subtracting $P(A \cap B)$ removes exactly the one extra count, leaving each outcome counted once. If $A$ and $B$ never overlap, $P(A \cap B) = 0$ and there is nothing to subtract — you get simple addition back.

How is the classical definition different from the axiomatic one

The axiomatic definition is the general rulebook: it says a probability must lie between 0 and 1, the sure event gets probability 1, and mutually exclusive events add. It works even when outcomes are not equally likely. The classical definition, $P(E) = n(E)/n(S)$ , is one concrete way to satisfy those rules when every outcome in a finite sample space is equally likely, as with a fair die or a well-shuffled deck. Class 11 mostly uses the classical definition because almost every problem involves equally likely outcomes.

Does this chapter cover probability with conditions, like drawing a second card given the first

No — conditional probability, independence, and probability distributions belong to Class XII. Class 11 ISC probability stays with single, unconditional events: sample spaces, mutually exclusive and exhaustive events, the classical definition, and the addition theorem. Problems here may still involve counting techniques from permutations and combinations, but never a "given that" clause.

Probability

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads