MathsISC Class 12 · Chapter 13

Probability ISC Class 12: Notes, Bayes' Theorem and Solved Questions

Updated 14 October 2026 · Fact-checked

Probability is Chapter 13 of ISC Class 12 Mathematics. It covers conditional probability, the addition and multiplication theorems, independent and dependent events, the theorem of total probability, Bayes' theorem, and random variables with their probability distributions and mean (expectation), with real-life problems throughout.

What Probability is about

Probability carries 9 of the 80 marks in ISC 2027, and Bayes' theorem regularly appears as a 5-mark real-life question in Section D.

Conditional probability: the probability of A given that B has happened is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , P(B) ≠ 0. Rearranged, this is the multiplication theorem, $P (A \cap B) = P (B) P (A ∣ B) = P (A) P (B ∣ A)$ , which handles draws without replacement. With the addition theorem, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , these let you combine events.

Independent events: A and B are independent if $P (A \cap B) = P (A) P (B)$ , i.e. knowing one does not change the probability of the other. Then their complements are independent too, and 'at least one' is best found as 1 − P(none). Mutually exclusive events (with non-zero probabilities) are never independent. Odds in favour a : b mean probability $\frac{a}{a + b}$ .

Total probability and Bayes' theorem: if $E_{1}, E_{2}, \dots, E_{n}$ partition the sample space (exactly one happens), then $P (A) = \sum P (E_{i}) P (A ∣ E_{i})$ , and Bayes' theorem reverses the conditioning: $P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum P ( E _{j} ) P ( A ∣ E _{j} )}$ . Draw a tree diagram to organise the numbers.

Random variables: a random variable assigns a number to each outcome; its probability distribution lists the values with probabilities (each between 0 and 1, summing to 1), and its mean (expected value) is $E (X) = \sum x_{i} p_{i}$ . Linear regression, once in the optional section, is no longer in the syllabus.

Weightage: Probability is part of the unit Probability, which carries 9 of the 80 marks in the ISC Mathematics. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Conditional probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ is the probability of A when B is known to have occurred: the sample space shrinks to B. Properties: P(B|B) = 1; $P (A^{'} ∣ B) = 1 - P (A ∣ B)$ ; for disjoint A, C: $P (A \cup C ∣ B) = P (A ∣ B) + P (C ∣ B)$ . In equally likely cases, $P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )}$ . In general P(A|B) ≠ P(B|A).; Worked example
If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, find P(A|B) and P(B|A).
Try it first, then show the solutionSolution
$P (A ∣ B) = \frac{0.3}{0.5} = 0.6$
$P (B ∣ A) = \frac{0.3}{0.6} = 0.5$
Answer: P(A|B) = 0.6, P(B|A) = 0.5 (here equal to P(A) and P(B): A and B happen to be independent)
Multiplication theorem; drawing without replacement: $P (A \cap B) = P (A) P (B ∣ A)$ ; for three events $P (A \cap B \cap C) = P (A) P (B ∣ A) P (C ∣ A \cap B)$ . Drawing without replacement changes the second probability because one item has gone; drawing with replacement keeps it the same. Selection problems can also be done with combinations: $\frac{^{5} C _{2}}{^{8} C _{2}}$ for two reds from 5 red and 3 blue.; Worked example
A bag has 5 red and 3 blue balls. Two are drawn one after the other without replacement. Find the probability that both are red.
Try it first, then show the solutionSolution
$P (R_{1}) = \frac{5}{8}$ ; then 4 red remain out of 7: $P (R_{2} ∣ R_{1}) = \frac{4}{7}$ .
$P (R_{1} \cap R_{2}) = \frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$ (check: $\frac{^{5} C _{2}}{^{8} C _{2}} = \frac{10}{28}$ ).
Answer: $\frac{5}{14}$
Independent events, 'at least one' and odds: A and B are independent if $P (A \cap B) = P (A) P (B)$ (equivalently P(A|B) = P(A)). Then A′ and B′, A and B′, A′ and B are also independent. For independent events, $P (at least one) = 1 - P (A^{'}) P (B^{'}) \dots$ . Mutually exclusive events (with non-zero probabilities) are dependent, since $P (A \cap B) = 0 \neq = P (A) P (B)$ . Odds in favour a : b give $P = \frac{a}{a + b}$ ; odds against a : b give $P = \frac{b}{a + b}$ .; Worked example
A and B are independent with P(A) = 0.4 and P(B) = 0.5. Find P(A ∪ B) in two ways.
Try it first, then show the solutionSolution
$P (A \cup B) = 0.4 + 0.5 - 0.4 \times 0.5 = 0.7$
Or $1 - P (A^{'}) P (B^{'}) = 1 - 0.6 \times 0.5 = 0.7$
Answer: 0.7
Theorem of total probability: If $E_{1}, \dots, E_{n}$ are mutually exclusive and exhaustive (a partition) with non-zero probabilities, then for any event A: $P (A) = P (E_{1}) P (A ∣ E_{1}) + \dots + P (E_{n}) P (A ∣ E_{n})$ . Think of a tree: first branch chooses the $E_{i}$ , second branch whether A happens; add the products along all paths ending in A.; Worked example
Machine M₁ makes 60% of a factory's output with 2% defective; M₂ makes 40% with 3% defective. Find the probability that a randomly chosen item is defective.
Try it first, then show the solutionSolution
P(D) = P(M₁)P(D|M₁) + P(M₂)P(D|M₂)
= 0.6 × 0.02 + 0.4 × 0.03 = 0.012 + 0.012 = 0.024
Answer: 0.024 (2.4%)
Bayes' theorem: Given that A has happened, the probability that it came from $E_{i}$ is $P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}$ . Steps: (1) name the events $E_{1}, E_{2}, \dots$ (the possible causes) and A (the observation); (2) write the prior probabilities $P (E_{i})$ and the likelihoods $P (A ∣ E_{i})$ ; (3) compute the denominator by total probability; (4) divide. A tree diagram prevents mistakes. When the evidence is 'failure' or 'defective', use the failure probabilities (1 − success rate).; Worked example
In the factory above, an item is found defective. Find the probability that it was made by M₁.
Try it first, then show the solutionSolution
P(D) = 0.024 (total probability).
$P (M_{1} ∣ D) = \frac{0.6 \times 0.02}{0.024} = \frac{0.012}{0.024} = 0.5$
Answer: 0.5
Random variables and probability distributions: A random variable X assigns a real number to each outcome of an experiment (number of heads, larger of two numbers drawn). Its probability distribution lists every value $x_{i}$ with $p_{i} = P (X = x_{i})$ , where $0 \leq p_{i} \leq 1$ and $\sum p_{i} = 1$ . If probabilities are given in terms of k, use $\sum p_{i} = 1$ to find k. Probabilities of ranges are sums: $P (X \leq 2) = p_{1} + p_{2}$ .; Worked example
X takes values 1, 2, 3, 4 with probabilities k, 2k, 3k, 4k. Find k and P(X ≥ 3).
Try it first, then show the solutionSolution
k + 2k + 3k + 4k = 1 ⇒ k = 0.1.
P(X ≥ 3) = 0.3 + 0.4 = 0.7.
Answer: k = 0.1; P(X ≥ 3) = 0.7
Mean (expectation) of a random variable: The mean or expected value is $E (X) = μ = \sum x_{i} p_{i}$ : the long-run average value per trial. It need not be a possible value of X. In games, E(gain) = Σ(gain × probability); a game is fair if the expected gain is 0. Do not average the probabilities or the values separately.; Worked example
Find the mean of X with values 1, 2, 3, 4 and probabilities 0.1, 0.2, 0.3, 0.4. In a game you win ₹10 with probability 0.2 and lose ₹2 otherwise: find your expected gain.
Try it first, then show the solutionSolution
E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.
E(gain) = 10(0.2) − 2(0.8) = 2 − 1.6 = 0.4.
Answer: E(X) = 3; expected gain ₹0.40 per game

Maths in real life: Probability

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.

Bayes' theorem in medical testing

Suppose 1% of people in a town have a certain infection. A test detects it in 95% of infected people but also gives a positive result for 5% of uninfected people.

A randomly chosen person tests positive. What is the probability that they actually have the infection?

Try it first, then show the solutionSolution

P(+) = 0.01 × 0.95 + 0.99 × 0.05 = 0.0095 + 0.0495 = 0.059
$P (infected ∣ +) = \frac{0.0095}{0.059} \approx 0.161$

Answer: Only about 16%: most positives are false alarms when the condition is rare

Memory hook: Rare cause + imperfect test = many false positives.

Independent events: at least one

Imagine you set two independent alarms for an early exam; each rings with probability 0.9.

What is the probability that at least one rings?

Try it first, then show the solutionSolution

P(neither) = 0.1 × 0.1 = 0.01
P(at least one) = 1 − 0.01 = 0.99

Answer: 0.99

Memory hook: At least one = 1 − none.

Expected value

Suppose a school fair sells raffle tickets at ₹50; one ticket in 40 wins a ₹1000 prize.

Find the expected gain from buying one ticket.

Try it first, then show the solutionSolution

Expected winnings = 1000 × (1/40) = ₹25
Expected gain = 25 − 50 = −25

Answer: −₹25 per ticket: on average you lose ₹25 (the fair raises money)

Memory hook: Expected value = Σ value × probability.

Topics in this chapter

Probability formula sheet

Conditional probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) \neq = 0$
Addition theorem: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Multiplication theorem: $P (A \cap B) = P (A) P (B ∣ A) = P (B) P (A ∣ B)$; Use for drawing without replacement.
Independence: $P (A \cap B) = P (A) P (B)$; Then complements are independent too.
At least one (independent): $P (at least one) = 1 - P (A^{'}) P (B^{'}) P (C^{'})$
Odds: $odds in favour a : b \Rightarrow P = \frac{a}{a + b}$; Odds against a : b ⇒ P = b/(a + b).
Total probability: $P (A) = \sum_{i = 1}^{n} P (E_{i}) P (A ∣ E_{i})$; $E_{i}$ partition the sample space.
Bayes' theorem: $P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}$
Probability distribution: $0 \leq p_{i} \leq 1, \sum p_{i} = 1$
Mean of a random variable: $E (X) = \sum x_{i} p_{i}$; Fair game: E(gain) = 0.

How the board asks Probability

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
One-mark items (Section A)	P(A\|B) from given values; E(X) from a table; odds; Statement or assertion items on independence and exclusivity.	Write the formula; for E(X) multiply and add.
Conditional and union (2 marks)	Find P(A ∪ B) from P(A), P(B) and P(A\|B); show events are (not) independent.	Get P(A ∩ B) first from the multiplication theorem.
Selections and 'at least'	Draws without replacement, committees, buying items from groups; at least one success.	Use combinations for selections; complement for 'at least one'; add mutually exclusive cases.
Total probability and Bayes (5 marks)	Factories, bags of balls, security agencies, medical tests.	Define events, tree diagram, total probability, then Bayes.
Distributions and mean	Construct a distribution (e.g. larger of two numbers, number of heads), find k, find the mean.	List outcomes systematically; check Σp = 1; E(X) = Σxp.

Tricks to remember Probability

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

Given goes downstairs
In P(A|B), the 'given' event B goes in the denominator.
Use it when: Writing conditional probability correctly.
Tree, then flip
Total probability walks down the tree; Bayes picks one branch and divides by the total.
Use it when: Bayes problems.
Exclusive is not independent
If two events can't happen together, knowing one changes the other's probability to 0.
Use it when: Assertion items.
At least one = 1 − none
For independent trials, compute the chance that all fail and subtract from 1.
Use it when: 'At least' questions (the error examiners report).
Odds a:b → a/(a + b)
Odds compare favourable to unfavourable; probability compares favourable to total.
Use it when: Odds questions.

Probability: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Conditional probability · 1 mark · MCQBoard level

If P(A ∩ B) = 0.2 and P(B) = 0.5, then P(A|B) is:
(a) 0.1
(b) 0.4
(c) 0.7
(d) 2.5

Try it first, then show the solutionSolution

$P (A ∣ B) = \frac{0.2}{0.5} = 0.4$ .

Answer: (b)

Example 2: Exclusive versus independent · 1 mark · Assertion-reasonBoard level

Assertion: Two mutually exclusive events A and B with P(A) > 0 and P(B) > 0 cannot be independent.
Reason: For such events P(A ∩ B) = 0, whereas P(A)P(B) > 0.
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Independence needs P(A ∩ B) = P(A)P(B); the reason shows this fails.

Answer: (a)

Example 3: Union from a conditional · 2 marks · Short answerHarder board level

If P(A) = 0.4, P(B) = 0.5 and P(A|B) = 0.3, find P(A ∪ B). [2]

Try it first, then show the solutionSolution

P(A ∩ B) = P(B)P(A|B) = 0.5 × 0.3 = 0.15.
P(A ∪ B) = 0.4 + 0.5 − 0.15 = 0.75.

Answer: 0.75

Example 4: At least one hit · 3 marks · Short answerHarder board level

Three archers hit a target with probabilities 0.5, 0.4 and 0.3, independently. Find the probability that the target is hit at least once, and that exactly one archer hits it. [3]

Try it first, then show the solutionSolution

P(no hits) = 0.5 × 0.6 × 0.7 = 0.21, so P(at least one) = 0.79.
Exactly one: 0.5 × 0.6 × 0.7 + 0.5 × 0.4 × 0.7 + 0.5 × 0.6 × 0.3 = 0.21 + 0.14 + 0.09.
= 0.44.

Answer: 0.79; 0.44

Example 5: Finding k and the mean · 3 marks · Short answerHarder board level

A random variable X takes the values 1, 2, 3, 4 with probabilities k, 2k, 3k, 4k. Find k, P(X ≤ 2) and E(X). [3]

Try it first, then show the solutionSolution

10k = 1 ⇒ k = 0.1.
P(X ≤ 2) = 0.1 + 0.2 = 0.3.
E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3.

Answer: k = 0.1; 0.3; 3

Example 6: Bayes' theorem with three bags · 5 marks · Long answerHarder board level

Bag I has 2 white and 3 red balls, Bag II has 4 white and 1 red, Bag III has 3 white and 3 red. A bag is chosen at random and a ball drawn from it is white.
(a) Find the probability of drawing a white ball. [2]
(b) Find the probability that it came from Bag II. [3]

Try it first, then show the solutionSolution

Events $E_{1}, E_{2}, E_{3}$ : bag chosen, each with probability 1/3. W: white drawn.
$P (W ∣ E_{1}) = \frac{2}{5}$ , $P (W ∣ E_{2}) = \frac{4}{5}$ , $P (W ∣ E_{3}) = \frac{3}{6} = \frac{1}{2}$ .
(a) $P (W) = \frac{1}{3} (\frac{2}{5} + \frac{4}{5} + \frac{1}{2}) = \frac{1}{3} \times \frac{17}{10} = \frac{17}{30}$ .
(b) $P (E_{2} ∣ W) = \frac{\frac{1}{3} \times \frac{4}{5}}{\frac{17}{30}} = \frac{8/30}{17/30}$ .
$= \frac{8}{17}$ .

Answer: (a) 17/30 (b) 8/17

Example 7: A positive test · 3 marks · Short answerHOTS / competency

A disease affects 1% of a population. A test is positive for 95% of those with the disease and for 5% of those without it. Find the probability that a person who tests positive has the disease. Comment on the result. [3]

Try it first, then show the solutionSolution

P(+) = 0.01(0.95) + 0.99(0.05) = 0.059.
P(D|+) = 0.0095/0.059 ≈ 0.161.
Because the disease is rare, false positives (0.0495) far outnumber true positives (0.0095); a second test is needed.

Answer: About 0.16

Example 8: Odds of independent events · 2 marks · Short answerHOTS / competency

The odds in favour of event A are 5 : 3 and the odds in favour of an independent event B are 2 : 3. Find the probability that both happen. [2]

Try it first, then show the solutionSolution

P(A) = 5/8, P(B) = 2/5.
P(A ∩ B) = (5/8)(2/5) = 1/4.

Answer: $\frac{1}{4}$

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 2 + 1 + 1 marks.

Defective bulbs from two suppliers

Suppose a shop buys 70% of its bulbs from supplier S₁ and 30% from supplier S₂. 5% of S₁'s bulbs and 10% of S₂'s bulbs are defective.

(i) · 1 mark
Draw a tree diagram and find the probability that a bulb is defective.
Answer
Answer: 0.065
0.7 × 0.05 + 0.3 × 0.10 = 0.035 + 0.030.
(ii) · 2 marks
A bulb is found defective. Find the probability that it came from S₂.
Answer
Answer: $\frac{6}{13} \approx 0.46$
Bayes: 0.030/0.065.
(iii) · 1 mark
Find the probability that a non-defective bulb came from S₁.
Answer
Answer: $\frac{0.665}{0.935} \approx 0.71$
P(S₁ ∩ D′) = 0.7 × 0.95 = 0.665; P(D′) = 1 − 0.065 = 0.935.
(iv) · 1 mark
Although S₂ supplies only 30% of the bulbs, it accounts for nearly half the defective ones. Explain.
Answer
Answer: Its defect rate is twice that of S₁.
0.3 × 0.10 = 0.030 is close to 0.7 × 0.05 = 0.035.

A spinner game

Suppose a game at a school fete uses a spinner. The prize X (₹) has the distribution: X = 0 with probability 0.5, X = 20 with probability 0.3, X = 50 with probability k. Each spin costs ₹20.

(i) · 1 mark
Find k.
Answer
Answer: 0.2
0.5 + 0.3 + k = 1.
(ii) · 1 mark
Find the expected prize.
Answer
Answer: ₹16
0 + 20(0.3) + 50(0.2) = 6 + 10.
(iii) · 2 marks
Is the game fair to the player? Find the expected gain per spin.
Answer
Answer: No; −₹4 per spin.
Expected gain = 16 − 20 = −4 ≠ 0.
(iv) · 1 mark
What should a spin cost for the game to be fair?
Answer
Answer: ₹16
Fair when cost = expected prize.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

If P(A) = 0.4, P(B) = 0.5 and P(A|B) = 0.3, find P(A ∪ B). (2 marks)

Step	Marks
P(A ∩ B) = P(B)P(A\|B) = 0.15	1
P(A ∪ B) = 0.75	1

X takes 1, 2, 3, 4 with probabilities k, 2k, 3k, 4k. Find k and E(X). (3 marks)

Step	Marks
Σp = 1 ⇒ k = 0.1	1
Writes E(X) = Σxp	1
E(X) = 3	1

Machines A, B, C make 50%, 30%, 20% of output with 1%, 2%, 3% defective. A defective item is found. Find the probability it was made by C. (5 marks)

Step	Marks
Events defined; priors 0.5, 0.3, 0.2	1
Likelihoods 0.01, 0.02, 0.03	1
P(D) = 0.005 + 0.006 + 0.006 = 0.017 (total probability)	1
Bayes: P(C\|D) = 0.006/0.017	1
= 6/17 ≈ 0.35	1

Where students lose marks in Probability

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

Mistake: Misreading 'at least one'
CISCE examiners' comments, 2024: candidates failed to see that 'at least one shot hits' is the complement of 'no hits'.
Wrong working
P(at least one) = 0.4 + 0.3 + 0.2 + 0.1 = 1.
Right working
1 − (0.6)(0.7)(0.8)(0.9) = 1 − 0.3024 = 0.6976.
Do this: For 'at least one' of independent events, subtract P(none) from 1.
Usually costs: 1 mark.
Mistake: Not recognising a Bayes problem
CISCE examiners' comments, 2024 and 2025: candidates did not recognise drawing without replacement, applied total probability but not Bayes' theorem, or did not see that failure rates were needed in a security-agency problem.
Wrong working
Stops at P(white and red) by total probability.
Right working
When the question asks 'given the result, which source?', divide the relevant branch by the total: Bayes.
Do this: Look for 'given that … find the probability that it came from …'.
Usually costs: 2-3 marks.
Mistake: Averaging instead of computing E(X)
CISCE examiners' comments, 2025: for the mean of a distribution, candidates took the average of the probabilities instead of Σx·P(X).
Wrong working
E(X) = (1/5 + 3/10 + 1/2)/3.
Right working
E(X) = Σxᵢpᵢ, e.g. 30(1/5) + 10(3/10) − 10(1/2) = 6 + 3 − 5 = 4.
Do this: Multiply each value by its probability, then add.
Usually costs: 1 mark.
Mistake: Treating dependent selections as independent
CISCE examiners' comments, 2025: in choosing three paintings from groups, candidates used independent-event products instead of combinations and multiplied when they should have added mutually exclusive cases; others could not convert odds into probabilities.
Wrong working
P(one from each) = (15/30)(5/30)(10/30).
Right working
$\frac{^{15} C _{1} ^{5} C _{1} ^{10} C _{1}}{^{30} C _{3}}$ ; for 'all from the same artist', add the three exclusive cases.
Do this: Selections from a fixed group use combinations; 'or' between exclusive cases means add.
Usually costs: 2 marks.

Practice questions on Probability

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 A and B are independent events, then P(A ∩ B) equals:
(a) P(A) + P(B)
(b) P(A)P(B)
(c) P(A|B)P(A)
(d) 0
Hint
Definition.
Answer
(b)
Q2 · 1 mark · MCQBoard level
X takes the values 0, 1, 2 with probabilities 0.3, 0.5, 0.2. E(X) is:
(a) 0.9
(b) 1
(c) 1.2
(d) 0.5
Hint
Σxp.
Answer
(a) 0.9
Q3 · 1 mark · MCQBoard level
If P(X = x) = kx for x = 1, 2, 3, 4, then k is:
(a) 1/4
(b) 1/10
(c) 1/5
(d) 1/8
Hint
Sum to 1.
Answer
(b) 1/10
Q4 · 1 mark · Very short answerBoard level
The odds against an event are 3 : 7. Write its probability.
Hint
Against means unfavourable first.
Answer
7/10
Q5 · 2 marks · Short answerBoard level
In a class, 40% of students study French, 30% study German and 10% study both. A student chosen studies German. Find the probability that they also study French.
Hint
P(F|G).
Answer
P(F|G) = 0.1/0.3 = 1/3
Q6 · 2 marks · Short answerHarder board level
Two balls are drawn without replacement from a bag of 5 red and 3 blue balls. Find the probability that both are red and that one of each colour is drawn.
Hint
Combinations or the multiplication theorem.
Answer
Both red 5/14; one of each $\frac{5 \times 3}{^{8} C _{2}} = \frac{15}{28}$ .
Q7 · 2 marks · Short answerHarder board level
A and B are independent with P(A) = 0.4 and P(B) = 0.5. Find P(A ∪ B) and P(exactly one of them).
Hint
Exactly one = A only + B only.
Answer
0.7; 0.4 × 0.5 + 0.6 × 0.5 = 0.2 + 0.3 = 0.5
Q8 · 2 marks · Short answerHOTS / competency
A die is thrown three times. Find the probability of getting at least one six.
Hint
Complement.
Answer
1 − (5/6)³ = 91/216
Q9 · 3 marks · Short answerHarder board level
Machine M₁ makes 60% of items (2% defective) and M₂ 40% (3% defective). Find the probability that a randomly chosen item is defective, and that a defective item came from M₂.
Hint
Total probability then Bayes.
Answer
P(D) = 0.024; P(M₂|D) = 0.012/0.024 = 0.5
Q10 · 3 marks · Short answerHarder board level
Three fair coins are tossed. Let X be the number of heads. Write the probability distribution of X and find its mean.
Hint
8 outcomes.
Answer
X = 0, 1, 2, 3 with probabilities 1/8, 3/8, 3/8, 1/8; E(X) = (0 + 3 + 6 + 3)/8 = 1.5.
Q11 · 3 marks · Short answerHarder board level
In a game you win ₹10 with probability 0.2 and otherwise lose ₹2. Find your expected gain and say whether the game favours you.
Hint
E = Σ gain × p.
Answer
E = 10(0.2) − 2(0.8) = ₹0.40 > 0: slightly in your favour.
Q12 · 5 marks · Long answerHarder board level
Three machines A, B, C produce 50%, 30% and 20% of a factory's items, with 1%, 2% and 3% defective respectively. An item chosen at random is defective. Find the probability that it was produced by (a) C, (b) A.
Hint
P(D) = 0.017.
Answer
P(D) = 0.005 + 0.006 + 0.006 = 0.017. (a) P(C|D) = 0.006/0.017 = 6/17 (b) P(A|D) = 0.005/0.017 = 5/17.
Q13 · 2 marks · Short answerHarder board level
Two cards are drawn without replacement from a pack of 52. Find the probability that both are aces.
Hint
(4/52)(3/51).
Answer
1/221

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.Formula for P(A|B)?Show answer
$\frac{P ( A \cap B )}{P ( B )}$ .
2.Multiplication theorem?Show answer
P(A ∩ B) = P(A)P(B|A).
3.Condition for independence?Show answer
P(A ∩ B) = P(A)P(B).
4.Can mutually exclusive events with non-zero probability be independent?Show answer
No.
5.Theorem of total probability?Show answer
P(A) = Σ P(Eᵢ)P(A|Eᵢ) for a partition {Eᵢ}.
6.Bayes' theorem?Show answer
$P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum P ( E _{j} ) P ( A ∣ E _{j} )}$ .
7.Conditions on a probability distribution?Show answer
Each pᵢ between 0 and 1; Σpᵢ = 1.
8.Mean of a random variable?Show answer
E(X) = Σxᵢpᵢ.

Best order to study it

Revise basic probability and the addition theorem.
Learn conditional probability and the multiplication theorem with draws without replacement.
Practise independent events, 'at least one' problems and odds.
Master total probability and Bayes' theorem with tree diagrams.
Practise constructing probability distributions and finding k.
Finish with means of random variables and fair games.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find P(A|B) and P(B|A) from given values?
Can you use the multiplication theorem for draws without replacement?
Can you test independence?
Can you find 'at least one' by the complement?
Can you convert odds into probabilities?
Can you set up a tree and apply total probability and Bayes?
Can you build a distribution and find k?
Can you compute E(X) and judge a game's fairness?

Probability: frequently asked questions

Is the binomial distribution in the ISC 2027 syllabus?

It is not named in the 2027 content, which lists conditional probability, the multiplication theorem, independent events, total probability, Bayes' theorem, random variables, probability distributions and the mean of a random variable.

Is linear regression still part of ISC Mathematics?

No. Linear regression was part of the old optional Section C and is not in the 2027 syllabus.

How do I recognise a Bayes' theorem question?

It gives several sources (machines, bags, agencies) with known proportions and success or failure rates, tells you an outcome has occurred, and asks which source it probably came from.

What is the difference between mutually exclusive and independent?

Mutually exclusive: the events cannot occur together (P(A ∩ B) = 0). Independent: one occurring does not change the other's probability (P(A ∩ B) = P(A)P(B)). Events with non-zero probability cannot be both.

How many marks does Probability carry?

9 of the 80 marks; expect Section A items plus a longer Bayes or distribution problem.

Do I need to draw a tree diagram?

It is not compulsory, but it makes Bayes and total probability questions much clearer and reduces errors.

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024