Probability Class 12: Notes, Methods and Solved Questions

What Probability is about

Weightage: This chapter is Unit VI, Probability, which carries 8 of the 80 theory marks.Source: CBSE Mathematics syllabus 2026-27, Unit VI Probability

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8 marks per paper. See the questions

Key concepts

Conditional probability

NCERT section 13.2

If E and F are events of the same experiment and $P(F)\ne 0$, the probability of E given that F has occurred is

$P(E|F)=\frac{P(E\cap F)}{P(F)}$

Think of F as the new, smaller sample space: only the outcomes of F are now possible, and we ask what fraction of them also lie in E. For equally likely outcomes, $P(E|F)=\frac{n(E\cap F)}{n(F)}$.

In general $P(E|F)\ne P(F|E)$: the condition matters.

Worked example

A card is drawn from a well-shuffled pack of 52. Find the probability that it is a king, given that it is a face card. [1 mark]

Try it first, then show the solutionSolution

1. F = face card: 12 cards. E ∩ F = kings: 4 cards.
2. $P(E|F)=\frac{4}{12}=\frac{1}{3}$

Answer: $\frac{1}{3}$

Properties of conditional probability

NCERT section 13.2.1

For events with $P(F)\ne 0$:

• $P(S|F)=P(F|F)=1$
• $P((A\cup B)|F)=P(A|F)+P(B|F)-P((A\cap B)|F)$; for disjoint A, B the last term is 0
• $P(E^{\prime }|F)=1-P(E|F)$

Conditional probabilities behave exactly like ordinary probabilities on the reduced sample space F, so all the Class 11 rules carry over. Remember also $0\le P(E|F)\le 1$.

Worked example

If P(A|F) = 0.4, P(B|F) = 0.5 and P((A ∩ B)|F) = 0.2, find P((A ∪ B)|F) and P(A′|F). [2 marks]

Try it first, then show the solutionSolution

1. P((A ∪ B)|F) = 0.4 + 0.5 − 0.2 = 0.7
2. P(A′|F) = 1 − 0.4 = 0.6

Answer: 0.7 and 0.6

Multiplication theorem on probability

NCERT section 13.3

From the definition of conditional probability:

$P(E\cap F)=P(E)P(F|E)=P(F)P(E|F)$

For three events: $P(E\cap F\cap G)=P(E)P(F|E)P(G|E\cap F)$.

Use it for draws without replacement: the probability for each draw is computed in the situation left by the earlier draws. With replacement, the draws are independent and the probabilities simply multiply.

Worked example

An urn has 5 red and 3 black balls. Two balls are drawn one after the other without replacement. Find the probability that both are red. [2 marks]

Try it first, then show the solutionSolution

1. P(first red) = $\frac{5}{8}$; P(second red | first red) = $\frac{4}{7}$
2. P(both red) = $\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}$

Answer: $\frac{5}{14}$

Independent events

NCERT section 13.4

E and F are independent if the occurrence of one does not affect the probability of the other: $P(E|F)=P(E)$ (with $P(F)\ne 0$). Equivalently,

$P(E\cap F)=P(E)P(F)$

Three events E, F, G are (mutually) independent if each pair satisfies this and also $P(E\cap F\cap G)=P(E)P(F)P(G)$.

Do not confuse independent with mutually exclusive: mutually exclusive events with non-zero probabilities can never be independent, because then $P(E\cap F)=0\ne P(E)P(F)$.

Worked example

A die is thrown. E: the number is a multiple of 3; F: the number is even. Are E and F independent? [2 marks]

Try it first, then show the solutionSolution

1. E = {3, 6}, F = {2, 4, 6}, E ∩ F = {6}
2. $P(E)P(F)=\frac{1}{3}\cdot \frac{1}{2}=\frac{1}{6}=P(E\cap F)$

Answer: Yes, they are independent.

Complements of independent events and "at least one"

NCERT section 13.4

Result: if E and F are independent, so are E and F′, E′ and F, and E′ and F′.

Proof (E and F′): $E=(E\cap F)\cup (E\cap F^{\prime })$, two disjoint events, so $P(E\cap F^{\prime })=P(E)-P(E\cap F)=P(E)-P(E)P(F)=P(E)[1-P(F)]=P(E)P(F^{\prime })$.

Hence for independent events, P(at least one occurs) $=1-P(E^{\prime })P(F^{\prime })$, and similarly for three or more events. This is the fastest way to handle "the problem is solved by at least one student" or "at least one alarm rings".

Worked example

Three students solve a problem independently with probabilities $\frac{1}{3}$, $\frac{1}{4}$ and $\frac{1}{5}$. Find the probability that the problem is solved. [2 marks]

Try it first, then show the solutionSolution

1. P(none solves) $=\frac{2}{3}\cdot \frac{3}{4}\cdot \frac{4}{5}=\frac{2}{5}$
2. P(solved) $=1-\frac{2}{5}=\frac{3}{5}$

Answer: $\frac{3}{5}$

Partition of a sample space and total probability

NCERT section 13.5.2

Events $E_1,E_2,\dots ,E_n$ form a partition of S if they are pairwise disjoint, their union is S, and each has non-zero probability: exactly one of them happens.

Theorem of total probability: for any event A,

$P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+\cdots +P(E_n)P(A|E_n)$

Proof sketch: $A=(A\cap E_1)\cup \cdots \cup (A\cap E_n)$, a disjoint union, and each $P(A\cap E_j)=P(E_j)P(A|E_j)$ by the multiplication rule. A tree diagram with one branch per cause makes the sum easy to see.

Worked example

Bag I has 3 red and 5 black balls; bag II has 4 red and 2 black balls. A fair coin decides the bag (heads: bag I). Find the probability that a ball drawn is red. [2 marks]

Try it first, then show the solutionSolution

1. $P(E_1)=P(E_2)=\frac{1}{2}$; $P(R|E_1)=\frac{3}{8}$, $P(R|E_2)=\frac{4}{6}=\frac{2}{3}$
2. $P(R)=\frac{1}{2}\cdot \frac{3}{8}+\frac{1}{2}\cdot \frac{2}{3}=\frac{3}{16}+\frac{1}{3}=\frac{25}{48}$

Answer: $\frac{25}{48}$

Bayes' theorem

NCERT section 13.5

If $E_1,\dots ,E_n$ partition S and $P(A)\ne 0$, then for each i,

$P(E_i|A)=\frac{P(E_i)P(A|E_i)}{{\sum }_{j=1}^{n}P(E_j)P(A|E_j)}$

Proof: $P(E_i|A)=\frac{P(E_i\cap A)}{P(A)}$; the numerator is $P(E_i)P(A|E_i)$ by the multiplication rule and the denominator is the total probability of A.

The $P(E_i)$ are prior probabilities (before the observation); the $P(E_i|A)$ are posterior probabilities (after observing A). Bayes' theorem answers "given the effect, which cause?"

Worked example

In the two-bag experiment above (coin decides the bag), the ball drawn is red. Find the probability that it came from bag I. [2 marks]

Try it first, then show the solutionSolution

1. $P(E_1|R)=\frac{P(E_1)P(R|E_1)}{P(R)}=\frac{3/16}{25/48}$
2. $=\frac{9}{48}\cdot \frac{48}{25}=\frac{9}{25}$

Answer: $\frac{9}{25}$

Bayes' theorem in testing problems

NCERT section 13.5

Medical and quality tests are a classic use of Bayes' theorem. Let D be "has the condition" and + be "tests positive". Then

$P(D|+)=\frac{P(D)P(+|D)}{P(D)P(+|D)+P(D^{\prime })P(+|D^{\prime })}$

When the condition is rare, even an accurate test gives a surprisingly low $P(D|+)$, because false positives from the large healthy group outnumber true positives. Always set up the two causes (D and D′) explicitly and use $P(D^{\prime })=1-P(D)$.

Worked example

A disease affects 1% of a population. A test is positive for 95% of people with the disease and for 2% of people without it. A person tests positive. Find the probability that the person has the disease. [3 marks]

Try it first, then show the solutionSolution

1. P(D) = 0.01, P(D′) = 0.99; P(+|D) = 0.95, P(+|D′) = 0.02
2. P(+) = 0.01 × 0.95 + 0.99 × 0.02 = 0.0095 + 0.0198 = 0.0293
3. $P(D|+)=\frac{0.0095}{0.0293}=\frac{95}{293}\approx 0.32$

Answer: $\frac{95}{293}$, only about 32%

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.

Topics in this chapter

• Conditional Probability
• Independent Events
• Bayes' Theorem

Probability formula sheet

Conditional probability

$P(E|F)=\frac{P(E\cap F)}{P(F)},P(F)\ne 0$

Complement under a condition

$P(E^{\prime }|F)=1-P(E|F)$

Union under a condition

$P((A\cup B)|F)=P(A|F)+P(B|F)-P((A\cap B)|F)$

Multiplication theorem

$P(E\cap F)=P(E)P(F|E)=P(F)P(E|F)$

Three events

$P(E\cap F\cap G)=P(E)P(F|E)P(G|E\cap F)$

Independent events

$P(E\cap F)=P(E)P(F)⟺P(E|F)=P(E)$

At least one (independent)

$P(\text{at least one})=1-P(E_1^{\prime })P(E_2^{\prime })\cdots P(E_n^{\prime })$

Exactly one of two

$P(\text{exactly one})=P(A)+P(B)-2P(A\cap B)$

Total probability

$P(A)={\sum }_{j=1}^{n}P(E_j)P(A|E_j)$

Bayes' theorem

$P(E_i|A)=\frac{P(E_i)P(A|E_i)}{{\sum }_{j=1}^{n}P(E_j)P(A|E_j)}$

How the board asks Probability

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

Tricks to remember Probability

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

• Given means shrink

  In P(E|F), F becomes the whole world: count only inside F.

  Use it when: Every conditional probability question.

• Independent multiplies, exclusive adds

  Independent: P(A ∩ B) = P(A)P(B). Mutually exclusive: P(A ∪ B) = P(A) + P(B) and P(A ∩ B) = 0.

  Use it when: Avoiding the classic mix-up.

• At least one = 1 − none

  For independent events, multiply the "fail" probabilities and subtract from 1.

  Use it when: "Problem solved", "target hit", "alarm rings".

• Tree first, Bayes second

  Draw a branch for each cause with its prior, then a branch for the outcome; total probability is the sum of the paths, Bayes is one path over the sum.

  Use it when: Total probability and Bayes questions.

• Prior times likelihood over evidence

  Bayes numerator = P(cause) × P(outcome | cause); denominator = P(outcome).

  Use it when: Writing Bayes' formula quickly.

Probability: board-level solved questions

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

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

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.

Where students lose marks in Probability

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

• ✗Mistake: Swapping P(A|B) and P(B|A)

  Dividing by the wrong event.

  Wrong working

  $P(A|B)=\frac{P(A\cap B)}{P(A)}$

  Right working

  $P(A|B)=\frac{P(A\cap B)}{P(B)}$

  ✓Do this: Divide by the probability of the event after the bar (the condition).

  Usually costs: The whole question.

• ✗Mistake: Treating mutually exclusive events as independent

  Writing P(A ∩ B) = P(A)P(B) for events that cannot happen together.

  ✓Do this: Mutually exclusive means P(A ∩ B) = 0; independent means P(A ∩ B) = P(A)P(B).

  Usually costs: 1-2 marks.

• ✗Mistake: Not defining events in Bayes questions

  Writing numbers without saying what E₁, E₂, A stand for.

  ✓Do this: Start with "Let E₁: …, E₂: …, A: …" and write each probability with its meaning.

  Usually costs: 1 mark in a 5-mark question.

• ✗Mistake: Forgetting to update after a draw without replacement

  Using the original counts for the second draw.

  ✓Do this: After each draw, reduce the total and the count of the colour drawn.

  Usually costs: 1-2 marks.

• ✗Mistake: Priors that do not add to 1

  Missing a cause or misreading percentages.

  ✓Do this: Check that P(E₁) + … + P(Eₙ) = 1 before using total probability.

  Usually costs: 2-3 marks.

NCERT exercise guide: Probability

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Probability: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8 marks per paper; 32 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Probability

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Probability question, by year

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.

1. Q1 · 1 mark · MCQBoard level

   If A and B are independent with P(A) = 0.4 and P(B) = 0.5, then P(A ∩ B) is:
   (a) 0.9
   (b) 0.2
   (c) 0.1
   (d) 0.45

   Hint

   Multiply.

   Answer

   (b) 0.2

2. Q2 · 1 mark · MCQBoard level

   If B ⊂ A and P(B) ≠ 0, then P(A|B) equals:
   (a) 0
   (b) P(A)
   (c) 1
   (d) P(B)

   Hint

   If B happens, A certainly happens.

   Answer

   (c) 1

3. Q3 · 1 mark · MCQBoard level

   Three fair coins are tossed. The probability of at least one head is:
   (a) $\frac{1}{8}$
   (b) $\frac{3}{8}$
   (c) $\frac{1}{2}$
   (d) $\frac{7}{8}$

   Hint

   1 − P(no head).

   Answer

   (d) $\frac{7}{8}$

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): When a die is thrown, the events "an even number" and "a number less than 3" are independent.
   Reason (R): Events E and F are independent if P(E ∩ F) = P(E)P(F).
   (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

   E ∩ F = {2}.

   Answer

   (a). $P(E)P(F)=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}=P(E\cap F)$.

5. Q5 · 2 marks · Very short answerHarder board level

   If P(A) = 0.6, P(B) = 0.3 and P(A ∪ B) = 0.7, find P(B|A) and P(A|B).

   Hint

   Find P(A ∩ B) first.

   Answer

   P(A ∩ B) = 0.6 + 0.3 − 0.7 = 0.2; $P(B|A)=\frac{0.2}{0.6}=\frac{1}{3}$, $P(A|B)=\frac{0.2}{0.3}=\frac{2}{3}$.

6. Q6 · 2 marks · Very short answerHarder board level

   Two cards are drawn one after the other without replacement from a pack of 52. Find the probability that both are spades.

   Hint

   Multiplication theorem.

   Answer

   $\frac{13}{52}\cdot \frac{12}{51}=\frac{1}{17}$

7. Q7 · 3 marks · Short answerHarder board level

   A and B are independent events with P(A) = 0.3 and P(A ∪ B) = 0.65. Find P(B) and P(A′ ∪ B′).

   Hint

   P(A ∪ B) = P(A) + P(B) − P(A)P(B).

   Answer

   Let P(B) = q: 0.3 + q − 0.3q = 0.65, so 0.7q = 0.35, q = 0.5. P(A′ ∪ B′) = 1 − P(A ∩ B) = 1 − 0.15 = 0.85.

8. Q8 · 3 marks · Short answerHarder board level

   In a city, 60% of commuters use buses and 40% use the metro. 10% of bus users and 5% of metro users arrive late. Find the probability that a commuter is late, and the probability that a late commuter used a bus.

   Hint

   Total probability, then Bayes.

   Answer

   P(late) = 0.6 × 0.1 + 0.4 × 0.05 = 0.08; P(bus | late) $=\frac{0.06}{0.08}=\frac{3}{4}$.

9. Q9 · 3 marks · Short answerHarder board level

   P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.1. Find the probability that exactly one of A and B occurs, and that neither occurs.

   Hint

   P(exactly one) = P(A) + P(B) − 2P(A ∩ B).

   Answer

   Exactly one: 0.5 + 0.4 − 0.2 = 0.7. Neither: 1 − P(A ∪ B) = 1 − 0.8 = 0.2.

10. Q10 · 5 marks · Long answerHarder board level

    A shop sells phones of brands X, Y and Z in the ratio 5 : 3 : 2. The probabilities of a fault within a year are 0.04, 0.06 and 0.1. Find (i) the probability that a phone sold is faulty, (ii) the probability that a faulty phone is of brand Z, (iii) the probability that a faulty phone is not of brand X.

    Hint

    Priors 0.5, 0.3, 0.2.

    Answer

    (i) 0.02 + 0.018 + 0.02 = 0.058. (ii) $\frac{0.02}{0.058}=\frac{10}{29}$. (iii) $1-\frac{0.02}{0.058}=\frac{19}{29}$.

11. Q11 · 3 marks · Short answerHOTS / competency

    If P(A) = $\frac{1}{2}$, P(B) = $\frac{1}{3}$ and P(A ∪ B) = $\frac{2}{3}$, are A and B independent? Mutually exclusive? Find P(A′|B).

    Hint

    Find P(A ∩ B).

    Answer

    $P(A\cap B)=\frac{1}{2}+\frac{1}{3}-\frac{2}{3}=\frac{1}{6}=P(A)P(B)$: independent, and not mutually exclusive since $P(A\cap B)\ne 0$. $P(A^{\prime }|B)=1-P(A|B)=1-\frac{1}{2}=\frac{1}{2}$.

12. Q12 · 3 marks · Short answerHOTS / competency

    A die is thrown twice. Given that the two numbers are different, find the probability that their sum is 6. Are the two events independent?

    Hint

    There are 30 ordered pairs with different numbers.

    Answer

    Sum 6 with different numbers: (1, 5), (5, 1), (2, 4), (4, 2), so $P=\frac{4}{30}=\frac{2}{15}$. P(sum 6) = $\frac{5}{36}$ ≠ $\frac{2}{15}$, so the events are not independent.

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. 1.Formula for P(E|F)?Show answer

   $\frac{P(E\cap F)}{P(F)}$, P(F) ≠ 0

2. 2.P(E′|F) in terms of P(E|F)?Show answer

   1 − P(E|F)

3. 3.Condition for independence of E and F?Show answer

   P(E ∩ F) = P(E)P(F)

4. 4.Can two mutually exclusive events with non-zero probabilities be independent?Show answer

   No.

5. 5.If E and F are independent, are E′ and F′ independent?Show answer

   Yes.

6. 6.What is a partition of a sample space?Show answer

   Pairwise disjoint events with non-zero probabilities whose union is the whole sample space.

7. 7.State the theorem of total probability.Show answer

   $P(A)=\sum P(E_j)P(A|E_j)$ for a partition $E_1,\dots ,E_n$.

8. 8.What does Bayes' theorem find?Show answer

   The probability of a cause given that an outcome has occurred (a posterior probability).

Quick revision checklist

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

• ☐Can you find P(E|F) by shrinking the sample space?
• ☐Can you use the properties P(E′|F) = 1 − P(E|F) and the conditional addition rule?
• ☐Can you find the probability of two or three draws without replacement?
• ☐Can you test whether two events are independent?
• ☐Can you prove that E and F′ are independent if E and F are?
• ☐Can you find P(at least one) for independent events?
• ☐Can you find unknown probabilities from conditions such as P(exactly one)?
• ☐Can you set up a partition and apply total probability?
• ☐Can you apply Bayes' theorem to a test with false positives?
• ☐Can you solve a ball-transfer problem with total probability and Bayes?

Probability: frequently asked questions