Probability notes: ISC Class 12 Mathematics

1. Laws of probability and the addition theorem

For events of a sample space $S$ : $0 \le P(A) \le 1$ , $P(S) = 1$ , $P(A') = 1 - P(A)$ .

Addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . For mutually exclusive events the overlap is empty and $P(A \cup B) = P(A) + P(B)$ .

The overlap is subtracted once because it was counted twice.
Law	Statement
Range	$0 \le P(A) \le 1$ ; $P(S) = 1$ , $P(\varnothing) = 0$
Complement	$P(A') = 1 - P(A)$
Addition theorem	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Mutually exclusive events	$P(A \cup B) = P(A) + P(B)$

Formulas

Addition theorem
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Key terms

Mutually exclusive events: Events that cannot happen together: their intersection is empty.
Complement: The event $A'$ that $A$ does not happen.

2. Conditional probability

If $P(B) > 0$ , the probability of $A$ given that $B$ has happened is $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ . With equally likely outcomes, $P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$ : restrict to the reduced sample space $B$ .

Properties: $P(B \mid B) = 1$ ; $P(A' \mid B) = 1 - P(A \mid B)$ ; the addition theorem holds given $B$ . In general $P(A \mid B) \ne P(B \mid A)$ .

In the imagined Dehradun survey, $P(\text{chess} \mid \text{science}) = \dfrac{3}{8}$ but $P(\text{science} \mid \text{chess}) = \dfrac{9}{16}$ .

Imagined survey data used throughout the chapter.
	Chess club	Not in chess club	Total
Science	$45$	$75$	$120$
Commerce	$35$	$45$	$80$
Total	$80$	$120$	$200$

The reduced sample space has 11 outcomes, not 36.
Outcomes with at least one 4	Sum 7?
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$	$(4, 3)$
$(1, 4), (2, 4), (3, 4), (5, 4), (6, 4)$	$(3, 4)$
$11$ outcomes in all	$2$ favourable: $\dfrac{2}{11}$

Formulas

Definition
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B) > 0$

Key terms

Conditional probability: The probability of an event when another event is known to have happened.
Reduced sample space: The outcomes of the given event, which become the new sample space.

3. Multiplication theorem and independence

Multiplication theorem: $P(A \cap B) = P(A)\,P(B \mid A)$ , and for three events $P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)$ . Use it for successive draws: without replacement the later factors change.

Independent events: $P(A \cap B) = P(A)\,P(B)$ , equivalently $P(A \mid B) = P(A)$ . If $A$ , $B$ are independent then so are $A$ and $B'$ , $A'$ and $B$ , $A'$ and $B'$ ; hence $P(\text{at least one}) = 1 - P(A')\,P(B')$ .

Events with non-zero probabilities cannot be both mutually exclusive and independent.

Both Darjeeling: $\frac{3}{10} \cdot \frac{2}{9} = \frac{1}{15}$.

Equal values: independent. Unequal: dependent.
Event B	$P(B)$	$P(A \cap B)$	$P(A)\,P(B)$	Independent?
Sum is 7	$\dfrac{1}{6}$	$\dfrac{1}{12}$	$\dfrac{1}{12}$	Yes
Sum is 8	$\dfrac{5}{36}$	$\dfrac{1}{12}$	$\dfrac{5}{72}$	No

Formulas

Multiplication theorem
$P(A \cap B) = P(A)\,P(B \mid A)$
Independence
$P(A \cap B) = P(A)\,P(B)$

Key terms

Independent events: Events for which the probability of both equals the product of their probabilities.
Dependent events: Events that are not independent: knowing one changes the chance of the other.

4. Theorem of total probability

If $E_1, \ldots, E_n$ form a partition of $S$ (pairwise disjoint, each $P(E_i) > 0$ , union $S$ ) and $A$ is any event, then

$P(A) = \sum_{i=1}^{n} P(E_i)\,P(A \mid E_i)$ .

In the imagined bolt factory: $P(D) = 0.5 \times 0.02 + 0.3 \times 0.03 + 0.2 \times 0.05 = 0.029$ .

$P(D) = 0.01 + 0.009 + 0.01 = 0.029$.

$A$ is cut into the pieces $A \cap E_1$, $A \cap E_2$, $A \cap E_3$.

Formulas

Total probability
$P(A) = \sum_{i=1}^{n} P(E_i)\,P(A \mid E_i)$

Key terms

Partition: Events that do not overlap, have non-zero probabilities and together make up the whole sample space.
Likelihood: $P(A \mid E_i)$ : the probability of the evidence under a given cause.

5. Bayes' theorem

With the same partition and $P(A) > 0$ :

$P(E_i \mid A) = \dfrac{P(E_i)\,P(A \mid E_i)}{\sum_j P(E_j)\,P(A \mid E_j)}$ .

$P(E_i)$ is the prior, $P(E_i \mid A)$ the posterior. The posteriors add to $1$ . A posterior exceeds its prior exactly when $P(A \mid E_i) > P(A)$ .

For a rare condition even an accurate test gives a modest posterior: in the imagined cattle test, $P(\text{sick} \mid +) = \dfrac{18}{67}$ .

A Bayes table keeps every step visible.
Machine	Prior	Likelihood	Product	Posterior
P	$0.5$	$0.02$	$0.010$	$\dfrac{10}{29}$
Q	$0.3$	$0.03$	$0.009$	$\dfrac{9}{29}$
R	$0.2$	$0.05$	$0.010$	$\dfrac{10}{29}$
Total	$1$		$0.029$	$1$

$P(\text{sick} \mid +) = \frac{18}{67}$: most positives are healthy animals.
1000 animals	Test positive	Test negative	Total
Sick	$18$	$2$	$20$
Healthy	$49$	$931$	$980$
Total	$67$	$933$	$1000$

Formulas

Bayes' theorem
$P(E_i \mid A) = \dfrac{P(E_i)\,P(A \mid E_i)}{\sum_j P(E_j)\,P(A \mid E_j)}$

Key terms

Prior probability: The probability of a cause before the evidence is seen.
Posterior probability: The probability of a cause after the evidence is seen.

6. Random variable, distribution and mean

A random variable $X$ assigns a real number to each outcome. Its probability distribution lists the values $x_i$ with probabilities $p_i$ , where $p_i \ge 0$ and $\sum p_i = 1$ .

The mean (expectation) is $E(X) = \sum x_i\,p_i$ , the long-run average value of $X$ .

Examples: fused bulbs, $E(X) = \dfrac{2}{3}$ ; difference of two dice, $E(X) = \dfrac{35}{18}$ ; $P(X = x) = k(x^2 + 1)$ for $x = 0, 1, 2, 3$ gives $k = \dfrac{1}{18}$ and $E(X) = \dfrac{7}{3}$ .

The probabilities add to 1.
$x$	$P(X = x)$
$0$	$\dfrac{2}{5}$
$1$	$\dfrac{8}{15}$
$2$	$\dfrac{1}{15}$

P(X = x)

$E(X) = \frac{70}{36} = \frac{35}{18}$.

Formulas

Mean of a random variable
$E(X) = \sum_{i} x_i\,p_i, \quad \sum_i p_i = 1$

Key terms

Random variable: A rule that gives a number to every outcome of an experiment.
Mean (expectation): $E(X) = \sum x_i\,p_i$ , the probability-weighted average of the values.

Exam tips

How ISC marks are awarded

Conditional probability and independence (2 marks): formula or reduced sample space, then the value; for independence show both sides of the product rule.
Total probability (2–3 marks): all causes with priors and likelihoods, then the sum.
Bayes (3–5 marks): total probability, the formula, the value, and an interpretation.
Distribution and mean (3–5 marks): each probability, the check that they add to $1$ , and the mean.

Common mistakes

Dividing by the wrong total, or confusing $P(A \mid B)$ with $P(B \mid A)$ .
Treating draws without replacement as independent.
Confusing mutually exclusive with independent.
Reporting the prior instead of the posterior.
Taking a simple average of the values instead of the weighted mean.

1. Laws of probability and the addition theorem

Formulas

Key terms

2. Conditional probability

Formulas

Key terms

3. Multiplication theorem and independence

Formulas

Key terms

4. Theorem of total probability

Formulas

Key terms

5. Bayes' theorem

Formulas

Key terms

6. Random variable, distribution and mean

Formulas

Key terms

Exam tips

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