MathsISC Class 12 · Probability

Conditional Probability, the Multiplication Theorem and Independent Events

Updated 15 October 2026 · Fact-checked

The conditional probability of A given B is $P(A|B) = \frac{P(A \cap B)}{P(B)}$, with $P(B) > 0$: the sample space shrinks to B. Rearranged, it gives the multiplication theorem $P(A \cap B) = P(B)P(A|B)$. Events A and B are independent when $P(A \cap B) = P(A)P(B)$, that is, when $P(A|B) = P(A)$.

Understand Conditional Probability and Independent Events

What "given" does. A conditional probability is a probability calculated after you have been told that some event $B$ has happened. Outcomes outside $B$ are now impossible, so $B$ becomes the new sample space and only the part of $A$ inside $B$ still counts:
$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) \neq = 0.$
With equally likely outcomes this is just counting inside $B$ : $P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )}$ .

Properties. $P (A ∣ B)$ behaves like an ordinary probability on the new sample space: $P (B ∣ B) = 1$ , $0 \leq P (A ∣ B) \leq 1$ , $P (A^{'} ∣ B) = 1 - P (A ∣ B)$ , and for disjoint $A$ and $C$ , $P (A \cup C ∣ B) = P (A ∣ B) + P (C ∣ B)$ . But in general $P (A ∣ B) \neq = P (B ∣ A)$ : they have the same numerator and different denominators. The ISC 2025 assertion-reason item turned on exactly this: $P (A ∣ B) = P (B ∣ A)$ forces $P (A) = P (B)$ (when $P (A \cap B) \neq = 0$ ), not that the events are mutually exclusive.

Multiplication theorem. Rearranging the definition,
$P (A \cap B) = P (A) P (B ∣ A) = P (B) P (A ∣ B) .$
For three events, $P (A \cap B \cap C) = P (A) P (B ∣ A) P (C ∣ A \cap B)$ . This is the rule for successive draws: each probability is worked out in the situation left by the earlier draws, so without replacement the denominators shrink.

Addition theorem. $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Questions often give $P (A ∣ B)$ and ask for $P (A \cup B)$ : first get $P (A \cap B) = P (B) P (A ∣ B)$ , then add. That was the 2-mark ISC 2024 question, where the examiners reported errors from not understanding conditional probability.

Independent events. $A$ and $B$ are independent when knowing that one happened does not change the chance of the other: $P (A ∣ B) = P (A)$ , equivalently $P (A \cap B) = P (A) P (B)$ . Then $A^{'}$ and $B$ , $A$ and $B^{'}$ , and $A^{'}$ and $B^{'}$ are independent too. Three events are mutually independent only if every pair and the triple satisfy the product rule.

Independent is not the same as mutually exclusive. Mutually exclusive means $P (A \cap B) = 0$ : if one happens the other cannot, which is the strongest possible dependence. Two events with non-zero probabilities cannot be both.

"At least one". For independent events, $P (at least one) = 1 - P (none) = 1 - P (A^{'}) P (B^{'}) P (C^{'}) \dots$ . The ISC 2024 examiners noted candidates who did not read "at least one" as the complement of "none".

Odds. Odds in favour of $E$ of $a : b$ mean $P (E) = \frac{a}{a + b}$ ; odds against $a : b$ mean $P (E) = \frac{b}{a + b}$ . In 2025 candidates lost a mark by not converting odds to probabilities.

Selections. When several items are taken at once from groups, count with combinations: $P = \frac{favourable selections}{all selections}$ . Do not multiply "independent" probabilities for items drawn from one pool.

Conditional Probability and Independent Events: formulas to remember

Conditional probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) > 0$; Equally likely outcomes: $\frac{n ( A \cap B )}{n ( B )}$ .
Complement given B: $P (A^{'} ∣ B) = 1 - P (A ∣ B)$; The condition stays the same.
Multiplication theorem: $P (A \cap B) = P (A) P (B ∣ A) = P (B) P (A ∣ B)$; Three events: $P (A) P (B ∣ A) P (C ∣ A \cap B)$ .
Addition theorem: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$; Use with the multiplication theorem to find $P (A \cup B)$ from $P (A ∣ B)$ .
Independence: $P (A \cap B) = P (A) P (B) ⟺ P (A ∣ B) = P (A)$; Complements of independent events are independent.
At least one (independent): $1 - \prod_{i} (1 - P (A_{i}))$; Complement of "none happens".
Odds: $in favour a : b \Rightarrow P = \frac{a}{a + b}$; Against $a : b$ gives $P = \frac{b}{a + b}$ .

How to solve Conditional Probability and Independent Events questions

Most conditional probability questions fall to the same five moves.

1Name the events in words and decide which one is the condition (the event after "given that" or "if it is known that").
2If outcomes are equally likely, list or count the reduced sample space $B$ and count $A \cap B$ inside it.
3Otherwise write $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ and find the numerator, often from the multiplication theorem.
4For successive draws, write each factor for the situation left by the previous draws (shrink the totals without replacement).
5For independent events, multiply; for "at least one", subtract "none" from 1.
6Check that each answer lies between 0 and 1 and that complementary answers add up to 1.

Conditional Probability and Independent Events in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Conditional probability from a two-way table

Suppose a school surveys 200 students: 120 travel by bus, and 45 of the bus users and 25 of the others arrive late at least once a week.

A student who arrives late at least once a week is chosen. Find the probability that the student travels by bus. Are "bus" and "late" independent?

Try it first, then show the solutionSolution

Late students: $45 + 25 = 70$ . Given late, the sample space is these 70.
$P (bus ∣ late) = \frac{45}{70} = \frac{9}{14}$ .
$P (bus) = \frac{120}{200} = \frac{3}{5}$ , which differs from $\frac{9}{14}$ , so the events are not independent.

Answer: $\frac{9}{14}$ ; not independent, since learning that a student was late raises the chance that they came by bus.

Memory hook: "Given late" means: throw away everyone who was never late, then count.

At least one, with independent events

Imagine a phone that tries three independent ways to get a signal: home Wi-Fi works with probability 0.7, mobile data with probability 0.8 and a neighbour's hotspot with probability 0.5.

Find the probability that the phone gets a signal by at least one route.

Try it first, then show the solutionSolution

All three fail: $0.3 \times 0.2 \times 0.5 = 0.03$ .
At least one works: $1 - 0.03 = 0.97$ .

Answer: 0.97.

Memory hook: At least one = 1 − all fail.

Tricks to remember Conditional Probability and Independent Events

Short memory aids. Say each one aloud once, then use it on a question the same day.

Given = new universe
Whatever follows "given" becomes the whole sample space; divide by its probability.
Use it when: Setting up $P (A ∣ B)$ .
Same top, different bottom
$P (A ∣ B)$ and $P (B ∣ A)$ share the numerator $P (A \cap B)$ ; only the denominators differ.
Use it when: Avoiding the swap error.
Exclusive is not independent
Exclusive: "if one, then not the other" (strongly dependent). Independent: "one tells you nothing about the other".
Use it when: Assertion-reason items.
At least one = 1 − none
Turn every "at least one" into one subtraction.
Use it when: Shooting, signals, machines.

Conditional Probability and Independent Events: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Reduced sample space · 1 mark · MCQBoard level

Two fair dice are thrown. Given that the sum is 8, the probability that at least one die shows 5 is:
(a) $\frac{1}{5}$
(b) $\frac{2}{5}$
(c) $\frac{1}{3}$
(d) $\frac{11}{36}$

Try it first, then show the solutionSolution

Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), five outcomes; two contain a 5.

Answer: (b) $\frac{2}{5}$ .

Example 2: From a conditional to a union · 2 marks · Short answerBoard level

If $P (A) = \frac{3}{8}$ , $P (B) = \frac{1}{2}$ and $P (A ∣ B) = \frac{1}{4}$ , find $P (A \cup B)$ and $P (B ∣ A)$ . (In the style of ISC 2024 Q6.)

Try it first, then show the solutionSolution

$P (A \cap B) = P (B) P (A ∣ B) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$ . (1 mark)
$P (A \cup B) = \frac{3}{8} + \frac{4}{8} - \frac{1}{8} = \frac{3}{4}$ ; $P (B ∣ A) = \frac{1/8}{3/8} = \frac{1}{3}$ . (1 mark)

Answer: $P (A \cup B) = \frac{3}{4}$ , $P (B ∣ A) = \frac{1}{3}$ .

Example 3: Three draws without replacement · 2 marks · Short answerHarder board level

A box holds 6 good and 4 faulty fuses. Three fuses are taken out one after another without replacement. Find the probability that the first two are good and the third is faulty.

Try it first, then show the solutionSolution

Multiplication theorem for three events: $\frac{6}{10} \cdot \frac{5}{9} \cdot \frac{4}{8}$ . (1 mark)
$= \frac{120}{720} = \frac{1}{6}$ . (1 mark)

Answer: $\frac{1}{6}$ .

Example 4: Testing independence · 3 marks · Short answerHarder board level

A card is drawn from a well-shuffled pack of 52. $A$ : the card is a king; $B$ : the card is a heart; $C$ : the card is a face card (J, Q, K) or an ace. Show that $A$ and $B$ are independent, and decide whether $A$ and $C$ are independent.

Try it first, then show the solutionSolution

$P (A) = \frac{4}{52} = \frac{1}{13}$ , $P (B) = \frac{1}{4}$ , $P (A \cap B) = \frac{1}{52} = P (A) P (B)$ : independent. (1 mark)
$P (C) = \frac{16}{52} = \frac{4}{13}$ ; $A \cap C = A$ , so $P (A \cap C) = \frac{1}{13}$ . (1 mark)
$P (A) P (C) = \frac{4}{169} \neq = \frac{1}{13}$ , so $A$ and $C$ are not independent (every king is in $C$ ). (1 mark)

Answer: $A$ , $B$ independent; $A$ , $C$ dependent.

Example 5: Selection from groups · 4 marks · Long answerHarder board level

Suppose a stall sells 12 mango, 8 litchi and 4 guava juice packs. A customer picks 3 packs at random. Find the probability that (i) all three are different flavours, (ii) all three are the same flavour. (In the style of ISC 2025 Q10.)

Try it first, then show the solutionSolution

Total selections: $(3 24) = 2024$ . (1 mark)
(i) $(1 12) (1 8) (1 4) = 384$ , so $P = \frac{384}{2024} = \frac{48}{253}$ . (1 mark)
(ii) $(3 12) + (3 8) + (3 4) = 220 + 56 + 4 = 280$ ; the three cases are mutually exclusive, so add. (1 mark)
$P = \frac{280}{2024} = \frac{35}{253}$ . (1 mark)

Answer: (i) $\frac{48}{253}$ ; (ii) $\frac{35}{253}$ .

Example 6: Odds and at least one · 4 marks · Case-basedHOTS / competency

Suppose a cricket match can be delayed by three independent causes: rain, bad light and a late team bus. The odds in favour of each cause occurring are 2 : 3, 3 : 5 and 1 : 4 respectively. Find the probability that (i) none of the causes occurs, (ii) exactly one occurs, (iii) the match is delayed (at least one occurs).

Try it first, then show the solutionSolution

Probabilities: $p_{1} = \frac{2}{5}$ , $p_{2} = \frac{3}{8}$ , $p_{3} = \frac{1}{5}$ ; complements $\frac{3}{5}, \frac{5}{8}, \frac{4}{5}$ .
(i) None: $\frac{3}{5} \cdot \frac{5}{8} \cdot \frac{4}{5} = \frac{3}{10}$ .
(ii) Exactly one: $\frac{2}{5} \cdot \frac{5}{8} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{3}{8} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{5}{8} \cdot \frac{1}{5} = \frac{40 + 36 + 15}{200} = \frac{91}{200}$ .
(iii) At least one: $1 - \frac{3}{10} = \frac{7}{10}$ .

Answer: (i) $\frac{3}{10}$ ; (ii) $\frac{91}{200}$ ; (iii) $\frac{7}{10}$ .

Where students lose marks in Conditional Probability and Independent Events

The slips examiners see most often, and what to write instead.

Mistake: Reading "at least one" wrongly
The ISC 2024 examiners reported candidates who did not treat "at least one hit" as the complement of "no hits".
Wrong working
Adding $0.4 + 0.3 + 0.2 + 0.1 = 1$ .
Right working
$1 - 0.6 \times 0.7 \times 0.8 \times 0.9 = 0.6976$ .
Do this: Write "at least one = 1 − none" before calculating.
Usually costs: 1 mark (the whole MCQ).
Mistake: Multiplying for a selection from one pool
In the 2025 paintings question, candidates treated choices from one group of 30 as independent events.
Do this: Selections made at once: use combinations over $(r n)$ .
Usually costs: 2 to 4 marks.
Mistake: Multiplying mutually exclusive cases
Also noted in 2025: "all same" is three separate cases that must be added.
Do this: "Or" between exclusive cases means add; "and" for one sequence means multiply.
Usually costs: 2 marks.
Mistake: Not converting odds
2025 candidates could not get a probability from odds of 5 : 2.
Do this: Favour $a : b$ means $\frac{a}{a + b}$ .
Usually costs: 1 mark.
Mistake: Treating exclusive as independent
Exclusive events have $P (A \cap B) = 0$ , not $P (A) P (B)$ .
Do this: Test each property separately using its definition.
Usually costs: 1 mark.

Exam tips

Write the defining formula first; it earns the method mark even if a number slips.
For dice, coins and cards, list the reduced sample space: it is quicker and safer than the formula.
Underline "without replacement", "at least" and "given that" as you read.
Leave answers as fractions in lowest terms unless decimals are asked for.
Prove independence by showing $P (A \cap B) = P (A) P (B)$ with numbers, not by saying "they do not affect each other".

Practice questions on Conditional Probability and Independent Events

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $P (B) = 0.4$ and $P (A \cap B) = 0.1$ , then $P (A^{'} ∣ B)$ is:
(a) 0.25
(b) 0.75
(c) 0.6
(d) 0.9
Hint
Find $P (A ∣ B)$ first.
Answer
(b) $1 - \frac{0.1}{0.4} = 0.75$ .
Q2 · 2 marks · Short answerBoard level
A family has two children. Given that at least one is a girl, find the probability that both are girls (boys and girls equally likely).
Hint
Reduced sample space: GG, GB, BG.
Answer
$\frac{1}{3}$ .
Q3 · 2 marks · Short answerHarder board level
If $A$ and $B$ are independent with $P (A) = 0.3$ and $P (A \cup B) = 0.65$ , find $P (B)$ .
Hint
$P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
Answer
$0.65 = 0.3 + 0.7 P (B)$ , so $P (B) = 0.5$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.2$ , then $A$ and $B$ are mutually exclusive.
Reason (R): For independent events, $P (A \cap B) = P (A) P (B)$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Is $P (A \cap B)$ zero?
Answer
(d): $P (A \cap B) = 0.2 \neq = 0$ , so not exclusive (they are in fact independent); R is a true statement.
Q5 · 2 marks · Short answerHarder board level
Suppose a quiz contestant answers each of 5 independent questions correctly with probability 0.6. Find the probability that she gets at least one wrong.
Hint
Complement of "all right".
Answer
$1 - 0. 6^{5} = 1 - 0.07776 = 0.92224$ .
Q6 · 3 marks · Short answerHarder board level
Two numbers are chosen one after the other without replacement from 1, 2, …, 9. Given that their sum is even, find the probability that both are odd.
Hint
Even sum: both odd or both even.
Answer
Both odd: $5 \cdot 4 = 20$ ordered pairs; both even: $4 \cdot 3 = 12$ . $P = \frac{20}{32} = \frac{5}{8}$ .
Q7 · 2 marks · Short answerHOTS / competency
Show that if $A$ and $B$ are independent, then $A^{'}$ and $B^{'}$ are independent.
Hint
De Morgan: $A^{'} \cap B^{'} = (A \cup B)^{'}$ .
Answer
$P (A^{'} \cap B^{'}) = 1 - P (A \cup B) = 1 - P (A) - P (B) + P (A) P (B) = (1 - P (A)) (1 - P (B)) = P (A^{'}) P (B^{'})$ .
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a security gate has two independent scanners. Scanner X flags a banned item with probability 0.9 and scanner Y with probability 0.8. A bag carrying a banned item passes through both.
(i) Find the probability that both flag it. (1 mark)
(ii) Find the probability that it is flagged at least once. (1 mark)
(iii) Given that it was flagged at least once, find the probability that X flagged it. (2 marks)
Hint
For (iii), X flagging is a subset of "at least once".
Answer
(i) 0.72. (ii) $1 - 0.1 \times 0.2 = 0.98$ . (iii) $\frac{0.9}{0.98} = \frac{45}{49}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define $P (A ∣ B)$ .Show answer
$\frac{P ( A \cap B )}{P ( B )}$ , $P (B) > 0$ .
2.State the multiplication theorem for two events.Show answer
$P (A \cap B) = P (A) P (B ∣ A)$ .
3.When are A and B independent?Show answer
When $P (A \cap B) = P (A) P (B)$ .
4.P(A) = 0.5, P(B) = 0.2, A and B independent: P(A ∪ B)?Show answer
$0.5 + 0.2 - 0.1 = 0.6$ .
5.Odds against an event are 7 : 3. What is its probability?Show answer
$\frac{3}{10}$ .
6.Can two mutually exclusive events with non-zero probabilities be independent?Show answer
No: $P (A \cap B) = 0$ but $P (A) P (B) > 0$ .

Conditional Probability and Independent Events: frequently asked questions

What is the difference between P(A|B) and P(A ∩ B)?

P(A \cap B) is the chance that both happen, measured against the whole sample space. P(A|B) is the chance of A measured only within B, after B is known to have happened. They are linked by P(A \cap B) = P(B)P(A|B).

How do I show that two events are independent?

Calculate P(A), P(B) and P(A \cap B) separately and show that the product equals the intersection. Saying that one event "does not affect" the other is not accepted as a proof.

When do I use combinations instead of multiplying probabilities?

Use combinations when several items are chosen together from a pool and order does not matter. Multiply conditional probabilities when items are drawn one after another. Both give the same answer when done correctly.

Is conditional probability needed for Bayes' theorem?

Yes. Bayes' theorem is built from the multiplication theorem and the theorem of total probability, so you need to be comfortable with P(A|B) first.

How many marks does this topic carry in ISC?

Probability carries 9 marks in the 2027 paper. Conditional probability appeared as a 2-mark question in 2024 and as an assertion-reason item and a 4-mark selection question in 2025.