Important Questions Class 12 Maths Chapter 13 Probability

Probability measures uncertainty by assigning numerical values to events in a random experiment.
Important Questions Class 12 Maths Chapter 13 help students practise conditional probability, independent events, Bayes theorem, random variables and binomial distribution.

Chapter 13 Probability is one of the most scoring parts of Class 12 Maths because most answers depend on identifying the correct condition first. A question may look like a simple dice, cards or balls problem, but one phrase like “given that”, “without replacement” or “found to be defective” changes the full solution method. This chapter trains students to move from basic probability to conditional probability, independent events, Bayes theorem, random variables and binomial distribution. 

For the 2026-27 exam, students should practise source-based questions like defective machines, medical tests, hostel records, truth-telling reports and repeated trials. These formats help build accuracy in Section A MCQs, Section D long answers and Section E case-study questions.

Key Takeaways

• Conditional Probability: P(E|F) = P(E ∩ F) / P(F), where P(F) is not zero.
• Multiplication Theorem: P(E ∩ F) = P(E) × P(F|E).
• Independent Events: Events E and F are independent when P(E ∩ F) = P(E) × P(F).
• Bayes Theorem: Bayes theorem finds the probability of a cause after an observed event.

Important Questions Class 12 Maths Chapter 13 Structure 2026-27

Section A: MCQs from Important Questions Class 12 Maths Chapter 13

Section A usually checks formulas, definitions, independence, conditional probability and quick Bayes theorem logic. Read the condition in the question before choosing the event.

Q1. If E and F are two events and P(F) is not zero, then P(E|F) is equal to:

1. P(E) / P(F)
   b. P(E ∩ F) / P(F)
   c. P(E ∪ F) / P(F)
   d. P(F) / P(E)

Answer: b. P(E ∩ F) / P(F)

Conditional probability means probability of E when F has already occurred.

Q2. If P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13, then P(A|B) is:

1. 4/9
   b. 9/4
   c. 7/9
   d. 4/13

Answer: a. 4/9

P(A|B) = P(A ∩ B) / P(B).

P(A|B) = (4/13) / (9/13) = 4/9.

Q3. If P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.2, then P(A|B) is:

1. 1/3
   b. 2/3
   c. 1/2
   d. 3/5

Answer: b. 2/3

P(A|B) = 0.2 / 0.3 = 2/3.

Q4. If P(B) = 0, then P(A|B) is:

1. 0
   b. 1
   c. Not defined
   d. P(A)

Answer: c. Not defined

Conditional probability P(A|B) needs P(B) to be non-zero.

Q5. Two events A and B are independent if:

1. P(A ∩ B) = 0
   b. P(A ∩ B) = P(A) × P(B)
   c. P(A ∪ B) = 1
   d. P(A) = P(B) always

Answer: b. P(A ∩ B) = P(A) × P(B)

This is the standard test for independent events.

Q6. If two events with non-zero probabilities are mutually exclusive, then they are:

1. Always independent
   b. Never independent
   c. Always equally likely
   d. Always complementary

Answer: b. Never independent

Mutually exclusive events cannot occur together, but independent events may occur together.

Q7. If A and B are independent, P(A) = 0.3 and P(B) = 0.4, then P(A ∩ B) is:

1. 0.7
   b. 0.12
   c. 0.1
   d. 0.4

Answer: b. 0.12

P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12.

Q8. If A and B are independent, P(A) = 0.3 and P(B) = 0.4, then P(A ∪ B) is:

1. 0.58
   b. 0.70
   c. 0.12
   d. 0.88

Answer: a. 0.58

P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

P(A ∪ B) = 0.3 + 0.4 - 0.12 = 0.58.

Q9. For independent events A and B, probability that at least one occurs is:

1. P(A)P(B)
   b. 1 - P(A′)P(B′)
   c. P(A′)P(B′)
   d. P(A) - P(B)

Answer: b. 1 - P(A′)P(B′)

At least one means complement of neither A nor B.

Q10. In Bayes theorem, the probabilities P(Ei) are called:

1. Posterior probabilities
   b. Priori probabilities
   c. Conditional outcomes
   d. Sample points

Answer: b. Priori probabilities

P(Ei) is the probability of a hypothesis before observing event A.

Q11. In Bayes theorem, P(Ei|A) is called:

1. Priori probability
   b. Posteriori probability
   c. Complement probability
   d. Impossible probability

Answer: b. Posteriori probability

Posteriori probability is the revised probability after event A occurs.

Q12. A set of events E1, E2, ..., En forms a partition of sample space S when the events are:

1. Equal only
   b. Pairwise disjoint, exhaustive and non-zero in probability
   c. Always independent
   d. Always mutually exclusive but not exhaustive

Answer: b. Pairwise disjoint, exhaustive and non-zero in probability

A partition divides the sample space into non-overlapping complete parts.

Q13. The theorem of total probability is used to find:

1. P(A), using a partition of sample space
   b. Only P(A ∩ B)
   c. Only P(A′)
   d. Only impossible events

Answer: a. P(A), using a partition of sample space

It combines probabilities across all possible causes or cases.

Q14. A random variable is a:

1. Real-valued function on a sample space
   b. Fixed constant only
   c. Set with no outcomes
   d. Graph only

Answer: a. Real-valued function on a sample space

A random variable assigns a real number to each outcome.

Q15. In a binomial distribution, each trial has:

1. Two possible outcomes
   b. Three possible outcomes
   c. No fixed probability
   d. Dependent outcomes only

Answer: a. Two possible outcomes

Binomial distribution uses success and failure outcomes.

Q16. If X follows binomial distribution with parameters n and p, then P(X = r) is:

1. nCr p^r q^(n-r)
   b. nPr p^r q^r
   c. p + q
   d. n + p

Answer: a. nCr p^r q^(n-r)

Here, q = 1 - p.

Q17. Assertion: If A and B are independent, then P(A|B) = P(A).

Reason: Occurrence of B does not affect the probability of A.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

This is the meaning of independence.

Q18. Assertion: Mutually exclusive events with non-zero probabilities cannot be independent.

Reason: For mutually exclusive events, P(A ∩ B) = 0.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

Independence would require P(A ∩ B) = P(A)P(B), which is non-zero.

Q19. If a die is thrown, event A = {3, 6} and event B = {2, 4, 6}. Then A and B are:

1. Dependent
   b. Independent
   c. Mutually exclusive
   d. Impossible

Answer: b. Independent

P(A) = 2/6 = 1/3.

P(B) = 3/6 = 1/2.

P(A ∩ B) = 1/6.

Since P(A) × P(B) = 1/6, events are independent.

Q20. If X is binomial with n = 5 and p = 1/2, then P(X = 2) is:

1. 5/32
   b. 10/32
   c. 2/5
   d. 1/10

Answer: b. 10/32

P(X = 2) = 5C2 × (1/2)^2 × (1/2)^3.

P(X = 2) = 10 × (1/2)^5 = 10/32.

Section B: Very Short Answer Questions from Class 12 Maths Chapter 13 Important Questions

Section B questions usually need one formula and direct substitution. Write the condition first when the question uses “given that”.

Q21. Define conditional probability class 12.

Conditional probability class 12 means the probability of one event when another event has already occurred.

If E and F are two events, then:

P(E|F) = P(E ∩ F) / P(F), where P(F) is not zero.

Q22. A family has two children. Find the probability that both are boys given that at least one is a boy.

Sample space after knowing at least one child is a boy:

{BB, BG, GB}

Favourable case for both boys:

{BB}

Required probability = 1/3.

Q23. If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A ∩ B).

Use multiplication theorem:

P(A ∩ B) = P(A) × P(B|A)

P(A ∩ B) = 0.8 × 0.4 = 0.32.

Q24. State multiplication theorem probability class 12.

Multiplication theorem probability class 12 states:

P(E ∩ F) = P(E) × P(F|E), where P(E) is not zero.

It can also be written as:

P(E ∩ F) = P(F) × P(E|F), where P(F) is not zero.

Q25. What are independent events class 12?

Independent events class 12 are events where occurrence of one event does not affect the probability of the other.

Events A and B are independent when:

P(A ∩ B) = P(A) × P(B).

Section C: Short Answer Questions from Probability Class 12 Important Questions

Section C questions usually test conditional probability, independence and small Bayes theorem applications. Keep sample spaces and favourable cases clear.

Q26. Ten cards numbered 1 to 10 are placed in a box. If the drawn card is known to be greater than 3, find the probability that it is even.

Let A be the event “card is even”.

A = {2, 4, 6, 8, 10}

Let B be the event “card is greater than 3”.

B = {4, 5, 6, 7, 8, 9, 10}

A ∩ B = {4, 6, 8, 10}

P(A|B) = Number of favourable outcomes in A ∩ B / Number of outcomes in B.

P(A|B) = 4 / 7.

So, the required probability is 4/7.

Q27. A die is thrown twice. Given that the sum is 6, find the probability that 4 appears at least once.

Let E be the event “4 appears at least once”.

Let F be the event “sum is 6”.

F = {(1,5), (2,4), (3,3), (4,2), (5,1)}

E ∩ F = {(2,4), (4,2)}

P(E|F) = Number of outcomes in E ∩ F / Number of outcomes in F.

P(E|F) = 2 / 5.

So, the required probability is 2/5.

Q28. Two cards are drawn without replacement from a pack of 52 cards. Find the probability that both cards are black.

There are 26 black cards in a pack of 52 cards.

Probability that first card is black = 26/52.

After drawing one black card, 25 black cards remain out of 51 cards.

Probability that second card is black given first is black = 25/51.

Required probability = 26/52 × 25/51.

Required probability = 1/2 × 25/51 = 25/102.

Q29. A box has 15 oranges, out of which 12 are good and 3 are bad. Three oranges are selected without replacement. Find the probability that all three are good.

Probability that first orange is good = 12/15.

Probability that second orange is good = 11/14.

Probability that third orange is good = 10/13.

Required probability = 12/15 × 11/14 × 10/13.

Cancel step:

12/15 = 4/5.

Required probability = 4/5 × 11/14 × 10/13.

Required probability = 44/91.

Q30. Check whether A and B are independent if P(A) = 1/2, P(B) = 1/2 and P(A ∩ B) = 1/4.

For independence:

P(A ∩ B) = P(A) × P(B)

P(A) × P(B) = 1/2 × 1/2 = 1/4.

Given P(A ∩ B) = 1/4.

So, A and B are independent events.

Q31. A bag I contains 3 red and 4 black balls. Bag II contains 5 red and 6 black balls. One bag is selected at random and one ball is drawn. If the ball is red, find the probability that it was drawn from Bag II.

Let E1 be the event of choosing Bag I.

Let E2 be the event of choosing Bag II.

P(E1) = 1/2 and P(E2) = 1/2.

Let A be the event “red ball is drawn”.

P(A|E1) = 3/7.

P(A|E2) = 5/11.

Using Bayes theorem:

P(E2|A) = [P(E2) × P(A|E2)] / [P(E1) × P(A|E1) + P(E2) × P(A|E2)]

P(E2|A) = [(1/2) × (5/11)] / [(1/2) × (3/7) + (1/2) × (5/11)]

Cancel 1/2 from numerator and denominator.

P(E2|A) = (5/11) / [(3/7) + (5/11)]

P(E2|A) = (5/11) / [(33 + 35) / 77]

P(E2|A) = (5/11) / (68/77)

P(E2|A) = 5/11 × 77/68 = 35/68.

So, the required probability is 35/68.

Section D: Long Answer Questions from Important Questions Class 12 Maths Chapter 13

Section D questions usually combine formulas with multiple steps. Write hypotheses, given probabilities and final substitution clearly.

Q32. A factory has machines A, B and C producing 25%, 35% and 40% of bolts. Their defective rates are 5%, 4% and 2%. A bolt is chosen and found defective. Find the probability that it was made by machine B.

Let B1, B2 and B3 be events that the bolt is made by machines A, B and C.

P(B1) = 0.25.

P(B2) = 0.35.

P(B3) = 0.40.

Let E be the event “bolt is defective”.

P(E|B1) = 0.05.

P(E|B2) = 0.04.

P(E|B3) = 0.02.

Using Bayes theorem:

P(B2|E) = [P(B2) × P(E|B2)] / [P(B1) × P(E|B1) + P(B2) × P(E|B2) + P(B3) × P(E|B3)]

Substitute values:

P(B2|E) = [0.35 × 0.04] / [(0.25 × 0.05) + (0.35 × 0.04) + (0.40 × 0.02)]

P(B2|E) = 0.014 / [0.0125 + 0.014 + 0.008]

P(B2|E) = 0.014 / 0.0345

P(B2|E) = 28/69.

So, the required probability is 28/69.

Q33. A doctor comes by train, bus, scooter or other means with probabilities 3/10, 1/5, 1/10 and 2/5. The probabilities of being late are 1/4, 1/3, 1/12 and 0 respectively. If the doctor is late, find the probability that he came by train.

Let T1, T2, T3 and T4 be events that the doctor comes by train, bus, scooter and other means.

P(T1) = 3/10.

P(T2) = 1/5.

P(T3) = 1/10.

P(T4) = 2/5.

Let E be the event “doctor is late”.

P(E|T1) = 1/4.

P(E|T2) = 1/3.

P(E|T3) = 1/12.

P(E|T4) = 0.

Using Bayes theorem:

P(T1|E) = [P(T1) × P(E|T1)] / [P(T1)P(E|T1) + P(T2)P(E|T2) + P(T3)P(E|T3) + P(T4)P(E|T4)]

Substitute values:

P(T1|E) = [(3/10) × (1/4)] / [(3/10 × 1/4) + (1/5 × 1/3) + (1/10 × 1/12) + (2/5 × 0)]

P(T1|E) = 3/40 / [3/40 + 1/15 + 1/120]

Convert to denominator 120:

3/40 = 9/120.

1/15 = 8/120.

1/120 = 1/120.

Denominator = 18/120.

P(T1|E) = 9/120 ÷ 18/120 = 1/2.

So, the required probability is 1/2.

Q34. A man speaks truth 3 out of 4 times. He throws a die and reports that it is six. Find the probability that it is actually six.

Let S1 be the event “six occurs”.

Let S2 be the event “six does not occur”.

P(S1) = 1/6.

P(S2) = 5/6.

Let E be the event “man reports six”.

If six actually occurs, he reports truth with probability 3/4.

P(E|S1) = 3/4.

If six does not occur, he reports six only when he lies.

P(E|S2) = 1/4.

Using Bayes theorem:

P(S1|E) = [P(S1) × P(E|S1)] / [P(S1)P(E|S1) + P(S2)P(E|S2)]

Substitute values:

P(S1|E) = [(1/6) × (3/4)] / [(1/6 × 3/4) + (5/6 × 1/4)]

P(S1|E) = 3/24 / [3/24 + 5/24]

P(S1|E) = 3/24 ÷ 8/24 = 3/8.

So, the probability that it is actually six is 3/8.

Q35. A question in a multiple-choice test is answered by a student. Probability that the student knows the answer is 3/4. Probability that he guesses is 1/4. If he guesses, probability of correct answer is 1/4. Find the probability that he knew the answer given that the answer is correct.

Let E1 be the event “student knows the answer”.

Let E2 be the event “student guesses the answer”.

P(E1) = 3/4.

P(E2) = 1/4.

Let A be the event “answer is correct”.

If the student knows the answer:

P(A|E1) = 1.

If the student guesses:

P(A|E2) = 1/4.

Using Bayes theorem:

P(E1|A) = [P(E1) × P(A|E1)] / [P(E1)P(A|E1) + P(E2)P(A|E2)]

Substitute values:

P(E1|A) = [(3/4) × 1] / [(3/4 × 1) + (1/4 × 1/4)]

P(E1|A) = 3/4 / [3/4 + 1/16]

Convert to denominator 16:

3/4 = 12/16.

Denominator = 12/16 + 1/16 = 13/16.

P(E1|A) = 12/16 ÷ 13/16 = 12/13.

So, the required probability is 12/13.

Section E: Case Study-Based Questions from Class 12 Maths Probability Important Questions

Section E questions usually give a real-life probability situation. Identify the events, write conditional probabilities and then apply the correct theorem.

Q36. Case Study: Medical test result

A laboratory blood test is 99% effective in detecting a disease when it is present. It gives a false positive result for 0.5% of healthy persons. In a population, 0.1% people actually have the disease. A person tests positive.

Q36(a). What is the probability that a person has the disease?

P(Disease) = 0.1% = 0.001.

Q36(b). What is the probability that a person is healthy?

P(Healthy) = 1 - 0.001 = 0.999.

Q36(c). What are the test probabilities?

P(Positive|Disease) = 0.99.

P(Positive|Healthy) = 0.005.

Q36(d). Find the probability that the person actually has the disease given that the test is positive.

Use Bayes theorem:

P(Disease|Positive) = [0.001 × 0.99] / [(0.001 × 0.99) + (0.999 × 0.005)]

Numerator = 0.00099.

Denominator = 0.00099 + 0.004995 = 0.005985.

P(Disease|Positive) = 0.00099 / 0.005985.

P(Disease|Positive) = 0.165 approximately.

So, the probability is about 0.165.

Q37. Case Study: Hostel and day scholars

In a college, 60% students live in hostel and 40% are day scholars. Previous results show that 30% hostel students and 20% day scholars attain A grade. One student is selected and found to have A grade.

Q37(a). Write the given probabilities.

Let H be the event “student is a hosteller”.

Let D be the event “student is a day scholar”.

P(H) = 0.60.

P(D) = 0.40.

Let A be the event “student gets A grade”.

P(A|H) = 0.30.

P(A|D) = 0.20.

Q37(b). Find the total probability of A grade.

P(A) = P(H)P(A|H) + P(D)P(A|D).

P(A) = 0.60 × 0.30 + 0.40 × 0.20.

P(A) = 0.18 + 0.08 = 0.26.

Q37(c). Find the probability that the A-grade student is a hosteller.

P(H|A) = [P(H)P(A|H)] / P(A).

P(H|A) = 0.18 / 0.26.

P(H|A) = 9/13.

Q37(d). Which theorem is used here?

Bayes theorem is used here.

It finds the probability of the student being a hosteller after knowing the student has A grade.

Q38. Case Study: Binomial distribution in a sample

Suppose 90% of people are right-handed. A random sample of 10 people is selected.

Q38(a). What is the distribution of the number of right-handed people?

The number of right-handed people follows a binomial distribution.

Here, n = 10 and p = 0.9.

Q38(b). What is q?

q = 1 - p.

q = 1 - 0.9 = 0.1.

Q38(c). Write the formula for getting exactly r right-handed people.

P(X = r) = 10Cr × (0.9)^r × (0.1)^(10-r).

Q38(d). Find the probability that all 10 people are right-handed.

P(X = 10) = 10C10 × (0.9)^10 × (0.1)^0.

P(X = 10) = (0.9)^10.

P(X = 10) = 0.3487 approximately.

Formula-Based Revision for Important Questions Class 12 Maths Chapter 13

Important questions class 12 maths chapter 13 should be revised through formulas and event language. Always define events before substitution.

Conditional Probability Class 12

P(E|F) = P(E ∩ F) / P(F)

Here:

P(E|F) means probability of E given F.

P(F) should not be zero.

Complement formula:

P(E′|F) = 1 - P(E|F)

Multiplication Theorem Probability Class 12

For two events:

P(E ∩ F) = P(E) × P(F|E)

Also:

P(E ∩ F) = P(F) × P(E|F)

For three events:

P(E ∩ F ∩ G) = P(E) × P(F|E) × P(G|E ∩ F)

Independent Events Class 12

A and B are independent if:

P(A ∩ B) = P(A) × P(B)

Also:

P(A|B) = P(A)

P(B|A) = P(B)

If A and B are independent, then A and B′ are also independent.

Theorem of Total Probability Class 12

If E1, E2, ..., En form a partition of sample space S, then:

P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + ... + P(En)P(A|En)

Use this when event A can happen through many possible cases.

Bayes Theorem Class 12

If E1, E2, ..., En form a partition of sample space S, then:

P(Ei|A) = [P(Ei)P(A|Ei)] / [P(E1)P(A|E1) + P(E2)P(A|E2) + ... + P(En)P(A|En)]

Use Bayes theorem when the result is known and the cause is asked.

Random Variable Class 12

A random variable is a real-valued function on the sample space.

Example:

When two coins are tossed, S = {HH, HT, TH, TT}.

If X is the number of heads:

X(HH) = 2.

X(HT) = 1.

X(TH) = 1.

X(TT) = 0.

Probability Distribution Class 12

A probability distribution lists the possible values of a random variable and their probabilities.

For a discrete random variable X:

Mean = Σ xi P(xi)

Variance = Σ xi² P(xi) - [Mean]²

Binomial Distribution Class 12

A binomial experiment has:

1. Fixed number of trials.
2. Two outcomes in each trial.
3. Constant probability of success.
4. Independent trials.

If X follows binomial distribution with parameters n and p:

P(X = r) = nCr p^r q^(n-r)

Here:

q = 1 - p

r = 0, 1, 2, ..., n

Mean = np.

Variance = npq.

Chapter-Wise Revision for Probability Class 12 Important Questions

Probability class 12 important questions should be revised in six parts: conditional probability, multiplication theorem, independent events, Bayes theorem, random variables and binomial distribution.

Start with conditional probability class 12. Most questions use the phrase “given that”, which means the sample space has changed.

Next, revise multiplication theorem probability class 12. It is used in card, ball and without-replacement questions.

Then revise independent events class 12. Check independence using P(A ∩ B) = P(A)P(B), not by guessing from the question.

After that, revise theorem of total probability class 12. It helps find the total chance of a result from different possible causes.

Next, practise Bayes theorem class 12. Use it when a result is known and the question asks which source or cause is most likely.

Finally, revise random variable class 12 and binomial distribution class 12. These questions need formula clarity and careful substitution.

Important Questions Class 12 Maths Chapter-Wise