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Probability

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Chapter Notes

Overview
This chapter builds on the classical and axiomatic probability studied in earlier classes, introducing the critical concept of conditional probability. It explores how the occurrence of one event can affect the probability of another, leading to the multiplication rule, the theorem of total probability, and Bayes' theorem. The chapter also formalizes the idea of independent events and introduces the concepts of random variables, their probability distributions, and the binomial distribution. The one big idea a student must walk away with is that probability is not just about counting outcomes, but about updating beliefs based on new information, which is the essence of conditional probability and Bayes' theorem.
Key Concepts (11)
Conditional Probability
The probability of an event E given that another event F has already occurred, denoted by P(E|F), is defined as P(E|F) = P(E ∩ F) / P(F), provided P(F) ≠ 0.
e.g. In a family with two children, the probability that both are boys given that at least one is a boy is 1/3, because the sample space reduces from {BB, BG, GB, GG} to {BB, BG, GB}.
Multiplication Theorem on Probability
For any two events E and F, the probability of their simultaneous occurrence is given by P(E ∩ F) = P(E) * P(F|E) = P(F) * P(E|F), provided P(E) and P(F) are non-zero.
e.g. Drawing two balls without replacement from an urn: P(first black and second black) = P(first black) * P(second black | first black).
Independent Events
Two events E and F are said to be independent if the occurrence of one does not affect the probability of the other, i.e., P(E|F) = P(E) and P(F|E) = P(F), or equivalently, P(E ∩ F) = P(E) * P(F).
e.g. Tossing a fair coin twice: the outcome of the first toss does not affect the outcome of the second toss.
Partition of a Sample Space
A set of events {E1, E2, ..., En} is said to represent a partition of the sample space S if they are pairwise disjoint (Ei ∩ Ej = φ for i ≠ j), exhaustive (E1 ∪ E2 ∪ ... ∪ En = S), and each has non-zero probability (P(Ei) > 0).
e.g. For any event E, the set {E, E'} forms a partition of the sample space.
Theorem of Total Probability
Let {E1, E2, ..., En} be a partition of the sample space S, and let A be any event. Then P(A) = Σ P(Ei) * P(A|Ei) for i = 1 to n.
e.g. If a construction job has a 0.65 chance of a strike, and the probability of completion on time is 0.32 with a strike and 0.80 without, then the total probability of completion on time is 0.65*0.32 + 0.35*0.80 = 0.488.
Bayes' Theorem
If {E1, E2, ..., En} is a partition of the sample space S and A is any event with P(A) > 0, then the posterior probability P(Ei|A) = [P(Ei) * P(A|Ei)] / Σ [P(Ej) * P(A|Ej)] for j = 1 to n.
e.g. If a red ball is drawn from one of two bags, the probability it came from Bag II is found using Bayes' theorem, given the prior probabilities of choosing each bag and the conditional probabilities of drawing a red ball from each.
Random Variable
A real-valued function defined on the sample space of a random experiment, whose value is determined by the outcome of the experiment.
e.g. If a coin is tossed three times, the random variable X could be the number of heads, taking values 0, 1, 2, or 3.
Probability Distribution
A listing of the possible values of a random variable along with their corresponding probabilities, such that the sum of all probabilities is 1.
e.g. For a random variable X representing the number of heads in two coin tosses, the distribution is P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4.
Mean of a Random Variable
The weighted average of the possible values of a random variable, where the weights are the probabilities. It is given by μ = Σ xi * P(X=xi).
e.g. For the number of heads in two coin tosses, the mean is 0*(1/4) + 1*(1/2) + 2*(1/4) = 1.
Variance of a Random Variable
A measure of the spread of a probability distribution, given by Var(X) = Σ (xi - μ)² * P(X=xi) = Σ xi² * P(X=xi) - μ².
e.g. For the number of heads in two coin tosses, the variance is (0-1)²*(1/4) + (1-1)²*(1/2) + (2-1)²*(1/4) = 0.5.
Binomial Distribution
A discrete probability distribution for the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success p. The probability of exactly x successes in n trials is P(X=x) = nCx * p^x * q^(n-x), where q = 1-p.
e.g. The probability of getting exactly 2 heads in 5 tosses of a fair coin is 5C2 * (1/2)² * (1/2)³ = 10/32 = 5/16.
Formulas & Equations (8)
P(E|F) = P(E ∩ F) / P(F)
Reads as: The probability of E given F equals the probability of E and F divided by the probability of F.
Variables: P(E|F): conditional probability of E given F (dimensionless); P(E ∩ F): probability of both E and F occurring; P(F): probability of F (non-zero).
Use when: When you need to find the probability of an event given that another event has already occurred.
⚠ Students often forget to check that P(F) ≠ 0, or they incorrectly write P(E|F) = P(F ∩ E) / P(E).
P(E ∩ F) = P(E) * P(F|E) = P(F) * P(E|F)
Reads as: The probability of E and F equals the probability of E times the probability of F given E, or the probability of F times the probability of E given F.
Variables: P(E ∩ F): probability of both events; P(E): probability of E; P(F|E): conditional probability of F given E; P(F): probability of F; P(E|F): conditional probability of E given F.
Use when: To find the probability of the intersection of two events, especially when events occur sequentially or conditionally.
⚠ Students apply this formula without checking if the events are dependent or independent, or they forget to use the conditional probability when events are not independent.
P(E ∩ F) = P(E) * P(F) (for independent events)
Reads as: The probability of E and F equals the probability of E times the probability of F.
Variables: P(E ∩ F): probability of both events; P(E): probability of E; P(F): probability of F.
Use when: Only when events E and F are known to be independent.
⚠ Students blindly multiply probabilities without verifying independence, or they assume mutually exclusive events are independent.
P(A) = Σ P(Ei) * P(A|Ei) for i = 1 to n
Reads as: The probability of A equals the sum over i of the probability of Ei times the probability of A given Ei.
Variables: P(A): total probability of event A; P(Ei): probability of event Ei (from a partition); P(A|Ei): conditional probability of A given Ei.
Use when: When an event A can occur through several mutually exclusive pathways (a partition of the sample space).
⚠ Students forget to ensure that the events Ei form a partition (exhaustive and mutually exclusive) before applying the formula.
P(Ei|A) = [P(Ei) * P(A|Ei)] / Σ [P(Ej) * P(A|Ej)] for j = 1 to n
Reads as: The probability of Ei given A equals the product of the prior probability of Ei and the likelihood of A given Ei, divided by the total probability of A.
Variables: P(Ei|A): posterior probability of Ei given A; P(Ei): prior probability of Ei; P(A|Ei): likelihood of A given Ei; denominator: total probability of A.
Use when: To find the probability of a 'cause' (Ei) given that an 'effect' (A) has occurred.
⚠ Students forget to compute the denominator correctly or confuse the prior and posterior probabilities.
Mean (μ) = Σ xi * P(X=xi)
Reads as: The mean equals the sum over all values of xi times the probability that X equals xi.
Variables: μ: mean of the random variable X; xi: possible values of X; P(X=xi): probability of X taking value xi.
Use when: To find the expected value or average outcome of a random variable.
⚠ Students forget to multiply each value by its probability, or they sum the values without weighting.
Variance (σ²) = Σ (xi - μ)² * P(X=xi) = Σ xi² * P(X=xi) - μ²
Reads as: The variance equals the sum of the squared deviations from the mean times their probabilities, or the mean of the squares minus the square of the mean.
Variables: σ²: variance of X; xi: possible values; μ: mean; P(X=xi): probability.
Use when: To measure the spread or dispersion of a probability distribution.
⚠ Students forget to subtract μ² when using the shortcut formula, or they use the wrong formula for standard deviation.
P(X=x) = nCx * p^x * q^(n-x), where q = 1-p
Reads as: The probability of exactly x successes in n trials is n choose x times p to the power x times q to the power n minus x.
Variables: n: number of trials; x: number of successes (0 ≤ x ≤ n); p: probability of success in a single trial; q: probability of failure (1-p).
Use when: When dealing with a fixed number of independent trials, each with two outcomes (success/failure) and constant probability of success.
⚠ Students forget to include the binomial coefficient nCx, or they confuse p and q (e.g., using p for failure).
How It Works
Calculating Conditional Probability
  1. Step 1: Identify the two events E and F from the problem.
  2. Step 2: Compute P(F), the probability of the event that is given (the condition).
  3. Step 3: Compute P(E ∩ F), the probability of both events occurring together.
  4. Step 4: Apply the formula P(E|F) = P(E ∩ F) / P(F).
💡 The sample space is effectively reduced to F, so only outcomes in F matter. Always ensure P(F) ≠ 0.
Applying the Multiplication Rule for Dependent Events
  1. Step 1: Identify the sequence of events (e.g., drawing cards without replacement).
  2. Step 2: Find the probability of the first event (e.g., P(first card is a king)).
  3. Step 3: Find the conditional probability of the second event given the first (e.g., P(second card is a king | first was a king)).
  4. Step 4: Multiply these probabilities: P(E ∩ F) = P(E) * P(F|E).
💡 The conditional probability changes because the sample space and the number of favorable outcomes change after the first event.
Checking for Independence of Two Events
  1. Step 1: Compute P(E), P(F), and P(E ∩ F).
  2. Step 2: Check if P(E ∩ F) = P(E) * P(F).
  3. Step 3: If true, events are independent; otherwise, they are dependent.
💡 Independence is not the same as mutual exclusivity. Mutually exclusive events with non-zero probabilities are always dependent.
Using Bayes' Theorem to Find Reverse Probability
  1. Step 1: Identify the partition of the sample space (the 'causes' or 'hypotheses') E1, E2, ..., En.
  2. Step 2: Write down the prior probabilities P(Ei) and the conditional probabilities P(A|Ei) for the event A.
  3. Step 3: Compute the total probability of A: P(A) = Σ P(Ei) * P(A|Ei).
  4. Step 4: Apply Bayes' theorem: P(Ei|A) = [P(Ei) * P(A|Ei)] / P(A).
💡 The denominator is always the total probability of the observed event A. The numerator is the joint probability of Ei and A.
Finding the Probability Distribution of a Random Variable
  1. Step 1: Define the random variable X and list all possible values it can take.
  2. Step 2: For each value, find the probability of the event that X equals that value.
  3. Step 3: Tabulate the values and their probabilities, ensuring the sum of all probabilities is 1.
💡 The random variable must be a real-valued function on the sample space. The sum of probabilities must always be exactly 1.
Solving a Binomial Distribution Problem
  1. Step 1: Verify that the experiment satisfies the conditions for a binomial experiment: fixed number of trials (n), independent trials, two outcomes (success/failure), constant probability of success (p).
  2. Step 2: Identify n, p, and the required number of successes x.
  3. Step 3: Apply the formula P(X=x) = nCx * p^x * (1-p)^(n-x).
💡 The trials must be independent, and the probability of success must remain the same for each trial. Do not forget the binomial coefficient.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Conditional Probability with Intersection Probability
✗ P(E|F) = P(E ∩ F)
→ Conditional probability is a ratio, not a joint probability. P(E|F) is the probability of E within the reduced sample space F, not the probability of both E and F in the original sample space.
✓ P(E|F) = P(E ∩ F) / P(F)
⚠ Assuming Independence Without Checking
✗ P(E ∩ F) = P(E) * P(F) for any two events E and F.
→ This formula is only valid for independent events. For dependent events, you must use the multiplication rule with conditional probability.
✓ First check if P(E ∩ F) = P(E) * P(F). If not, use P(E ∩ F) = P(E) * P(F|E).
⚠ Misapplying Bayes' Theorem Denominator
✗ P(E1|A) = [P(E1) * P(A|E1)] / [P(E1) * P(A|E1)]
→ The denominator must be the total probability of A, which is the sum over all partitions, not just the numerator.
✓ P(E1|A) = [P(E1) * P(A|E1)] / [P(E1)*P(A|E1) + P(E2)*P(A|E2) + ...]
⚠ Forgetting the Binomial Coefficient in Binomial Distribution
✗ P(X=x) = p^x * q^(n-x)
→ This gives the probability of a specific sequence of successes and failures, not the probability of exactly x successes in any order. The binomial coefficient accounts for the number of ways to arrange the successes.
✓ P(X=x) = nCx * p^x * q^(n-x)
⚠ Treating 'At Least One' Problems Incorrectly
✗ P(at least one success) = p + p + ... (adding probabilities)
→ Events are not mutually exclusive, so adding probabilities leads to double-counting. The correct approach uses the complement.
✓ P(at least one success) = 1 - P(no success) = 1 - q^n
Compare & Contrast
Definition and relationship
Independent Events
Mutually Exclusive Events
Independent events can occur together (P(E∩F) = P(E)P(F) ≠ 0), while mutually exclusive events cannot occur together (P(E∩F) = 0). Two mutually exclusive events with non-zero probabilities are always dependent.
Interpretation and formula
Conditional Probability P(E|F)
Joint Probability P(E ∩ F)
P(E|F) is the probability of E given that F has occurred, calculated as P(E∩F)/P(F). P(E∩F) is the probability that both E and F occur simultaneously, without any condition. They are equal only if P(F)=1.
Exam Focus
Marks pattern: Typically 1-2 MCQs (1 mark each), 1 short answer (2-3 marks) on conditional probability or independence, 1 long answer (4-5 marks) on Bayes' theorem or binomial distribution, and sometimes a 1-mark question on properties of probability.
Most asked: Conditional probability problems (especially with dice, cards, and family children) · Bayes' theorem (especially with bags and balls, or test reliability) · Independent events and multiplication rule · Binomial distribution (finding probabilities, mean, and variance)
Don't skip: Bayes' theorem problems with 3 or more partitions (e.g., machines producing bolts, bags with balls); Binomial distribution: finding P(X=x), mean, and variance; Proving independence or dependence of events using P(E∩F) = P(E)P(F)
One-Line Revision (read aloud before the exam)
  1. Conditional probability P(E|F) = P(E∩F)/P(F) reduces the sample space to F.
  2. Multiplication rule: P(E∩F) = P(E) * P(F|E) = P(F) * P(E|F).
  3. Events E and F are independent if P(E∩F) = P(E) * P(F).
  4. Independent events are not the same as mutually exclusive events.
  5. A partition of a sample space is a set of pairwise disjoint, exhaustive events with non-zero probabilities.
  6. Theorem of total probability: P(A) = Σ P(Ei) * P(A|Ei) for a partition {Ei}.
  7. Bayes' theorem finds P(Ei|A) = [P(Ei) * P(A|Ei)] / P(A).
  8. A random variable is a real-valued function on a sample space.
  9. The mean of a random variable is μ = Σ xi * P(X=xi).
  10. The variance of a random variable is σ² = Σ (xi-μ)² * P(X=xi) = Σ xi²*P(X=xi) - μ².
  11. Binomial distribution: P(X=x) = nCx * p^x * q^(n-x), where q = 1-p.
  12. For a binomial distribution, mean = np and variance = npq.
  13. The probability of at least one success in n trials is 1 - q^n.
  14. In Bayes' theorem, prior probabilities are updated to posterior probabilities after observing an event.

MCQs — 10 Questions with Answers

Q1. If P(A) = 3/7, P(B) = 5/7, and P(A∩B) = 2/7, then P(A|B) = ?

  • A. 2/5 ✓
  • B. 3/5
  • C. 5/7
  • D. 2/7

Answer: A — Using P(A|B) = P(A∩B)/P(B) = (2/7)/(5/7) = 2/5.

Q2. A family has two children. What is the probability both are boys given at least one is a boy?

  • A. 1/2 ✓
  • B. 1/3
  • C. 1/4
  • D. 2/3

Answer: A — Sample space with at least one boy: {(b,b), (g,b), (b,g)}, favorable outcome for both boys: {(b,b)}, so P = 1/3, not 1/2 (common error: P(b,b) from total space is 1/4).

Q3. If P(E|F) = 0.6 and P(E'|F) = x, then x = ?

  • A. 0.6
  • B. 0.4 ✓
  • C. 0.3
  • D. Cannot be determined

Answer: B — By Property 3, P(E'|F) = 1 − P(E|F) = 1 − 0.6 = 0.4.

Q4. Ten numbered cards 1–10 are in a box. If drawn card is known to be > 5, probability it is even is?

  • A. 1/5
  • B. 2/5
  • C. 3/5 ✓
  • D. 3/4

Answer: C — Given event F: {6,7,8,9,10}, favorable event E∩F: {6,8,10}, so P(E|F) = 3/5.

Q5. Which statement about conditional probability is NOT correct?

  • A. P(E|F) requires P(F) > 0
  • B. P(E|F) = P(F|E) always ✓
  • C. P(E|F) + P(E'|F) = 1
  • D. P(S|F) = 1

Answer: B — P(E|F) ≠ P(F|E) in general; they are different conditional probabilities representing different ordered events.

Q6. Two events A and B satisfy: P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2. Then P(A|B) and P(B|A) are respectively:

  • A. 0.5 and 0.4 ✓
  • B. 0.2 and 0.2
  • C. 0.5 and 0.4
  • D. 0.4 and 0.5

Answer: A — P(A|B) = 0.2/0.4 = 0.5; P(B|A) = 0.2/0.5 = 0.4; they are unequal, illustrating order-dependence.

Q7. For mutually exclusive events A, B and event F, if P(F) ≠ 0, then P((A∪B)|F) equals:

  • A. P(A|F) + P(B|F) − P((A∩B)|F)
  • B. P(A|F) + P(B|F) ✓
  • C. P(A|F) × P(B|F)
  • D. P(A|F) − P(B|F)

Answer: B — For mutually exclusive events, P((A∩B)|F) = 0, so Property 2 simplifies to P((A∪B)|F) = P(A|F) + P(B|F).

Q8. A box contains 5 red and 3 blue balls. One red ball is drawn and not replaced. Probability next ball is red given first was red:

  • A. 5/8
  • B. 4/7 ✓
  • C. 4/8
  • D. 5/7

Answer: B — After drawing one red ball without replacement, 4 red and 3 blue remain out of 7 total, so P(second red | first red) = 4/7.

Q9. **Assertion**: P(E|F) = P(E∩F)/P(F) is valid for all events F. **Reason**: This formula defines conditional probability when P(F) = 0.

  • A. Both assertion and reason are true; reason explains assertion
  • B. Both are true; reason does not explain assertion
  • C. Assertion is true; reason is false ✓
  • D. Assertion is false; reason is true

Answer: C — The assertion is true and valid only when P(F) ≠ 0; the reason is false because the formula is not defined when P(F) = 0.

Q10. In 100 students, 60 like Math and 45 like Science, 25 like both. If a student likes Math, probability they like Science is:

  • A. 45/100
  • B. 25/45
  • C. 25/60 ✓
  • D. 60/100

Answer: C — P(Science|Math) = P(both)/P(Math) = 25/60, using students liking Math as the reduced sample space.

Flashcards

What is conditional probability P(E|F)?

P(E|F) = P(E∩F)/P(F), the probability of event E occurring given that event F has already occurred, provided P(F) ≠ 0.

Why must P(F) ≠ 0 in conditional probability formula?

Because division by zero is undefined, and an event with zero probability (impossible event) cannot serve as a given condition to reduce the sample space.

State Property 1 of conditional probability.

P(S|F) = P(F|F) = 1, meaning the probability of the entire sample space or the event itself given that event has occurred is always 1.

State Property 3 of conditional probability.

P(E'|F) = 1 − P(E|F), the complement rule holds: conditional probability of complement equals 1 minus conditional probability of the event.

For disjoint events A and B, what is P((A∪B)|F)?

P((A∪B)|F) = P(A|F) + P(B|F), since the intersection P((A∩B)|F) = 0 for disjoint events.

Two cards drawn without replacement: first is red. How does this affect probability second is red?

The reduced sample space now has one fewer red card and one fewer total card, making P(second red | first red) = (red cards−1)/(total cards−1), which differs from initial probability.

In a conditional probability problem, what does E∩F represent?

E∩F represents outcomes where both events E and F occur simultaneously; these are the favorable outcomes for calculating P(E|F).

A die shows number > 3. What is sample space for conditional probability?

The reduced sample space is {4, 5, 6} instead of {1, 2, 3, 4, 5, 6}, containing only outcomes satisfying the given condition.

Prove: P(E|F) + P(E'|F) = 1

Since E and E' are disjoint and E∪E' = S, we have P((E∪E')|F) = P(E|F) + P(E'|F) = P(S|F) = 1 by Property 1.

If P(A|B) = P(A), what can you conclude about events A and B?

Events A and B are independent because knowing B occurred does not change the probability of A, meaning P(A∩B) = P(A)·P(B).

Important Board Questions

Define conditional probability and state the condition under which P(E|F) = P(E∩F)/P(F) is valid. Give one example. [2 marks]

State P(E|F) formula explicitly, require P(F) ≠ 0, provide any real example like coin/card/dice with two events and calculate using the formula.

Prove that P(E|F) + P(E'|F) = 1. Then use this to find P(E'|F) if P(E|F) = 7/13. [5 marks]

Start with P(S|F) = P(E∪E'|F) = 1 from Property 1; expand using Property 2 for disjoint events E and E'; substitute values to compute P(E'|F) = 6/13.

A die is thrown twice. Event A: sum of numbers is 8; Event B: first throw is at least 3. Find P(A|B) and verify using definition. [6 marks]

List all outcomes in B: (3,1) to (6,6) — 24 outcomes; find A∩B: {(3,5), (4,4), (5,3)} — 3 outcomes; calculate P(A|B) = 3/24 = 1/8; verify by computing P(A∩B)/P(B) = (3/36)/(24/36) = 3/24.

Assertion-Reason Questions for Probability

Q1. Assertion (A):

If P(A ∩ B) = P(A) × P(B), then events A and B are independent.

Reason (R):

Independence means P(A|B) = P(A), which is equivalent to the given condition.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The multiplication rule for independent events directly gives the equivalence: P(A|B) = P(A ∩ B)/P(B) = P(A) × P(B)/P(B) = P(A). The reason correctly explains why the assertion holds.

Q2. Assertion (A):

For any events A and B with P(B) ≠ 0, P(A|B) is always less than or equal to P(A).

Reason (R):

Conditional probability reduces the sample space, which always decreases the probability.

Answer: D — A is false but R is true — The assertion is false. Counter-example: A = {3, 4, 5, 6}, B = {4, 5, 6} in a die roll. P(A) = 2/3, but P(A|B) = 3/3 = 1, which is greater. The reason is also misleading because conditioning can increase probability. Both are false.

Q3. Assertion (A):

P(E'|F) = 1 - P(E|F) holds for all events E and F where P(F) ≠ 0.

Reason (R):

This follows from the fact that E and E' partition the sample space, and their conditional probabilities must sum to 1.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true and directly related. Since S = E ∪ E' (disjoint union), P(S|F) = P(E ∪ E'|F) = P(E|F) + P(E'|F) = 1 (Property 1). This correctly justifies the assertion.

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Case-Based Questions (CBQ) for Probability

Case Study:

A mobile phone company in India tracks network usage patterns. Among its 10,000 customers, 6,000 subscribe to data plans (D), 4,000 subscribe to voice plans (V), and 2,500 have both. The company randomly selects a customer who uses voice plans to offer a special data upgrade. Among customers with voice plans, the company knows that 60% are in urban areas (U) and 40% are in rural areas.

Q1. What is the probability that a randomly selected voice plan customer also has a data plan?

  • A. 2500/6000
  • B. 2500/4000 ✓
  • C. 2500/10000
  • D. 6000/10000

Answer: B

Q2. Among the 4,000 voice plan customers, 60% are urban. How many voice plan customers are urban?

  • A. 2400 ✓
  • B. 1600
  • C. 2500
  • D. 3600

Answer: A

Q3. If a voice plan customer is selected at random, what is the probability they have both plans given they are from an urban area? (Assume among 2400 urban voice customers, 1500 have data plans.)

  • A. 1500/2400 = 0.625 ✓
  • B. 1500/4000 = 0.375
  • C. 1500/6000 = 0.25
  • D. 2500/2400 > 1 (invalid)

Answer: A

Q4. What is the probability that a randomly selected voice customer is rural AND has both plans if 1000 of the 1600 rural voice customers have data plans?

  • A. 1000/10000 = 0.1
  • B. 1000/4000 = 0.25 ✓
  • C. 1600/4000 = 0.4
  • D. 2500/6000 ≈ 0.42

Answer: B

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Previous Year Questions (PYQ) for Probability

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

Assertion (A): In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is 2/3. Reason (R): For any two events A and B, P(A|B) = P(A ∩ B)/P(B)

  • Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true and Reason (R) is false.
  • Assertion (A) is false and Reason (R) is true.

CBSE 2026 · Set 65/1/1 · 3 marks

Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.

CBSE 2026 · Set 65/1/1 · 1 mark

In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each. A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold thereby raising a fund of ₹ 1,00,00,000. Rohan bought one ticket.

In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each. A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold thereby raising a fund of ₹ 1,00,00,000. Rohan bought one ticket. What are the possible amounts, the person can win?

CBSE 2026 · Set 65/1/2 · 1 mark

Assertion (A): In an experiment of throwing an unbiased die, the probability of getting a prime number given that the number appearing on the die being odd is 2/3. Reason (R): For any two events A and B, P(A|B) = P(A ∩ B)/P(B).

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A)
  • Assertion (A) is true and Reason (R) is false
  • Assertion (A) is false and Reason (R) is true

CBSE 2026 · Set 65/1/2 · 1 mark

In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each. A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold there by raising a fund of ₹ 1,00,00,000. Rohan bought one ticket.

What are the possible amounts, the person can win?

CBSE 2026 · Set 65/2/1 · 1 mark

For two events A and B such that P(A) ≠ 0 and P(B) ≠ 1, P(A'|B) =

  • 1 - P(A|B)
  • 1 - P(A'|B)
  • [1 - P(A∩B)]/P(B)
  • [1 - P(A'∩B)]/P(B)

CBSE 2026 · Set 65/2/1 · 3 marks

The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed. The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that (i) target is hit (ii) atleast one shot misses the target.

CBSE 2026 · Set 65/2/1 · 1 mark

Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems.

What is the probability that selected person is a female?

CBSE 2026 · Set 65/2/2 · 1 mark

For two events A and B such that P(A) ≠ 0 and P(B) ≠ 1, P(A'/B) =

  • 1 – P(A/B)
  • 1 – P(A'/B)
  • 1 – P(A ∩ B)/P(B)
  • 1 – P(A ∪ B)/P(B)

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

If events A and B are such that P(A|B) = P(B|A), then :

  • A ⊂ B but A ≠ B
  • A = B
  • A ∩ B = ∅
  • P(A) = P(B)

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

The probability of getting an odd prime number on each die, when a pair of dice is rolled, is :

  • 1/18
  • 1/9
  • 1/36
  • 1/6

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

In a family with three children, the probability that the eldest child is a boy, given that the family has at least one boy, is :

  • 1/2
  • 1/3
  • 2/3
  • 4/7

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