Class 9 Probability Previous Year Questions (2020–2025): 13 Solved PYQs + Board Exam Strategy

Most students approach Probability by re-reading the chapter: What is a random experiment? What are mutually exclusive events? But after the third read, the definitions blur and don't stick during exams. Past papers flip this: you encounter a real question, struggle, look up the concept, and cement it in context. CBSE Probability questions are remarkably repetitive—the same core ideas (sample space, event probability, axiomatic rules) reappear with different numbers. By solving 5–10 PYQs, you'll recognise 80% of exam variants instantly. Research on spaced repetition confirms that retrieval practice (solving questions) beats passive review by 3× in long-term retention. For Chapter 14 specifically, you need to master: (1) listing sample spaces for dice/coin experiments, (2) calculating P(E) = n(E)/n(S), (3) applying rules P(E ∪ F) = P(E) + P(F) − P(E ∩ F), and (4) identifying mutually exclusive and independent events. Past papers force you to do all four under timed pressure, which is exactly what your 2-hour board exam demands. Start with 1-mark questions to warm up your intuition, then jump to 5-mark problems that combine multiple concepts.

1-mark questions test definitional clarity and basic recall. Here are the five most common patterns: **Question 1:** A fair die is rolled once. What is the probability of getting a prime number? **Answer:** Prime numbers on a die: 2, 3, 5 (total 3). Sample space size: 6. P(prime) = 3/6 = 1/2. **Question 2:** Two coins are tossed simultaneously. What is the probability of getting at most one head? **Answer:** Sample space: {HH, HT, TH, TT} (size 4). Favorable outcomes: {HT, TH, TT} (size 3). P(at most 1 head) = 3/4. **Question 3:** A card is drawn from a standard deck of 52 cards. What is the probability it is a king? **Answer:** Number of kings = 4. Total cards = 52. P(king) = 4/52 = 1/13. **Question 4:** Are the events 'rolling an odd number' and 'rolling an even number' on a die mutually exclusive? Why? **Answer:** Yes. Because a single roll cannot be both odd and even simultaneously; P(odd ∩ even) = 0, so they cannot occur together. **Question 5:** If P(E) = 0.65, what is P(not E)? **Answer:** P(E') = 1 − P(E) = 1 − 0.65 = 0.35. This uses the complement rule from axiomatic probability.

3-mark questions demand two steps: identify the sample space correctly, then apply a probability formula or rule. You'll lose marks for incomplete reasoning, so show your work clearly. **Question 1:** A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. (a) Write the sample space. (b) Find P(red or blue). (c) Are these events mutually exclusive? **Solution:** (a) Sample space S = {5 red, 3 blue, 2 green}, n(S) = 10. (b) n(red) = 5, n(blue) = 3. Events 'red' and 'blue' are mutually exclusive, so P(red ∪ blue) = P(red) + P(blue) = 5/10 + 3/10 = 8/10 = 4/5. (c) Yes, because a single draw cannot yield both red and blue simultaneously. **Question 2:** Two dice are rolled together. Find the probability that (i) the sum is 7, (ii) both numbers are the same, (iii) the product is at least 12. **Solution:** n(S) = 36. (i) Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, so P = 6/36 = 1/6. (ii) Same: {(1,1), (2,2), ..., (6,6)}, so P = 6/36 = 1/6. (iii) Product ≥ 12: {(2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}, so P = 17/36. **Question 3:** A card is drawn from a shuffled deck. Let E = event of drawing a spade, F = event of drawing a face card (J, Q, K). Are E and F independent? Justify. **Solution:** P(E) = 13/52 = 1/4. P(F) = 12/52 = 3/13. P(E ∩ F) = 3/52 (3 face cards are spades). For independence, P(E ∩ F) should equal P(E) × P(F). Check: (1/4) × (3/13) = 3/52. Yes, they are independent. **Question 4:** Three coins are tossed. Find P(at least 2 heads). **Solution:** S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8. At least 2 heads: {HHH, HHT, HTH, THH}, count = 4. P = 4/8 = 1/2. **Question 5:** A spinner has 8 equal sectors: 3 red, 2 blue, 2 yellow, 1 green. It is spun once. Find (i) P(red), (ii) P(not blue), (iii) P(red or yellow). **Solution:** (i) P(red) = 3/8. (ii) P(not blue) = 1 − 2/8 = 6/8 = 3/4. (iii) P(red ∪ yellow) = 3/8 + 2/8 = 5/8 (mutually exclusive).

5-mark questions combine multiple concepts: constructing a sample space, applying multiple rules, and interpreting results. These often involve conditional probability, compound events, or real-world scenarios. Full solution structure is essential to avoid losing marks for incomplete reasoning. **Question 1:** A bag contains 4 white and 6 black balls. Two balls are drawn one after another without replacement. (a) Draw a tree diagram. (b) Find the probability that both balls are white. (c) Find the probability that one is white and one is black. (d) Verify that the sum of all probabilities = 1. **Solution:** (a) Tree diagram: First draw—P(W) = 4/10, P(B) = 6/10. After removing one ball, second draw branches depend on first. (b) P(WW) = (4/10) × (3/9) = 12/90 = 2/15. (c) P(one W, one B) = P(WB) + P(BW) = (4/10)(6/9) + (6/10)(4/9) = 24/90 + 24/90 = 48/90 = 8/15. (d) P(WW) + P(one W, one B) + P(BB) = 2/15 + 8/15 + (6/10)(5/9) = 2/15 + 8/15 + 30/90 = 12/90 + 48/90 + 30/90 = 90/90 = 1. ✓ **Question 2:** A fair die is rolled twice. Let E = 'sum is even', F = 'product is even'. (a) Find P(E) and P(F). (b) Find P(E ∩ F). (c) Are E and F mutually exclusive? Are they independent? Justify both. **Solution:** n(S) = 36. (a) Sum is even when both are odd or both are even: {odd, odd} has 3×3 = 9 outcomes; {even, even} has 3×3 = 9 outcomes. So n(E) = 18, P(E) = 18/36 = 1/2. Product is even when at least one is even: complement is 'both odd' = 9 outcomes. So n(F) = 27, P(F) = 27/36 = 3/4. (b) E ∩ F = both even (all products are even). n(E ∩ F) = 9, so P(E ∩ F) = 9/36 = 1/4. (c) P(E ∩ F) = 1/4 ≠ 0, so NOT mutually exclusive. For independence, check if P(E ∩ F) = P(E) × P(F): (1/2)(3/4) = 3/8. But P(E ∩ F) = 1/4. Since 1/4 ≠ 3/8, they are NOT independent. **Question 3:** A committee of 3 is chosen randomly from 4 men and 3 women. (a) Find the sample space size (total committees). (b) Find the probability that the committee has at least 1 woman. (c) Find the probability that the committee has more women than men. (d) List the condition(s) under which these two events are mutually exclusive. **Solution:** (a) n(S) = C(7,3) = 35 committees. (b) Complement: all men = C(4,3) = 4. So P(at least 1 woman) = 1 − 4/35 = 31/35. (c) More women than men with 3 people: 2W+1M or 3W+0M. n(2W,1M) = C(3,2) × C(4,1) = 3 × 4 = 12. n(3W,0M) = C(3,3) = 1. Total = 13. P(more W than M) = 13/35. (d) Events 'at least 1 woman' and 'more women than men' are NOT mutually exclusive because '3 women' satisfies both. They overlap.

The 2024–25 rationalized CBSE syllabus slightly reordered Probability content but retained core topics: random experiments, events, and axiomatic probability. However, based on the emphasis on competency-based assessment, examiners are shifting toward: (1) **Real-world application scenarios** instead of abstract die/coin problems. Expect questions like 'A survey shows 40% of students prefer online classes. If 5 students are sampled...' This tests whether you can translate word problems into sample spaces. (2) **Multi-step reasoning over single-formula answers.** A 3-mark question now often asks 'Why is this event mutually exclusive?' rather than just 'Find P(E)'. You must justify your reasoning using definitions. (3) **Less memorization of complex formulas.** The conditional probability formula P(E|F) = P(E ∩ F)/P(F) is not explicitly tested in Class 9 but provides context for independence. Focus on understanding rather than formula recall. (4) **Emphasis on verification.** Questions now ask you to verify that probabilities sum to 1 or check if two events are independent by calculation. This shifts marks toward process, not just final answers. At cbsetutor.ai, we've updated our question bank to reflect these shifts: every question includes justification prompts and real-world framing. To stay ahead, practice questions that ask 'Why?' not just 'Find', and always show your sample space construction explicitly.

Time is the enemy in board exams. Here's a battle-tested strategy for maximizing marks in Probability: **Before the exam (1 week prior):** Solve at least 8 full PYQs under timed conditions (50 min per paper). Identify which question types trip you: do you struggle with tree diagrams, sample space enumeration, or applying the addition rule? Target those weaknesses. Memorize exactly four things: (1) Sample space definition and notation. (2) P(E) = n(E)/n(S) formula. (3) P(E ∪ F) = P(E) + P(F) − P(E ∩ F). (4) Definitions of mutually exclusive (E ∩ F = ∅) and independent (P(E ∩ F) = P(E)P(F)). Everything else derives from these. **During the exam:** (a) **Read the question twice.** The first read clarifies context; the second identifies what's being asked. (b) **Write your sample space explicitly,** even for 1-mark questions. Examiners award partial credit for correct methodology. (c) **Identify the event clearly.** Circle or underline which outcomes are favorable. (d) **Choose formulas last.** First understand the problem, then decide if you need P(E), P(E ∪ F), or verification of independence. (e) **Avoid shortcuts in 5-mark questions.** A full tree diagram or list of outcomes takes 30 seconds but secures 2 marks. Skipping it and jumping to the answer costs 1–2 marks if your calculation is wrong. (f) **Check your probability value.** Is it between 0 and 1? If not, you've made an error. (g) **Allocate time:** 1-mark questions: 1.5 min max. 3-mark: 4 min. 5-mark: 6 min. If you're stuck, move on and return later. **Common pitfall to avoid:** Confusing 'at least k' with 'exactly k'. 'At least 2 heads in 3 coins' means 2 or 3 heads, not just 2. Always enumerate or use complement carefully.

Probability in Class 9 is foundational. Once you've mastered this chapter, you're ready for: **Class 10 Statistics & Probability:** Class 10 deepens conditional probability and Bayes' theorem (not in Class 9), introduces mean, mode, and median of grouped data. Your solid understanding of sample spaces and events makes Class 10 transitions smooth. **Class 11 Permutations & Combinations:** The binomial coefficient C(n,r) appears in Class 11 and is essential for calculating n(E) and n(S) in complex scenarios. Chapter 14 sets this foundation. **Real-world applications:** Probability underpins epidemiology (COVID models), finance (risk assessment), and machine learning (Bayesian classifiers). Understanding events and axiomatic rules now prepares you for advanced streams later. Start a 3-day free trial at cbsetutor.ai to access 25+ video solutions for Chapter 14, instant doubt-resolution with AI tutors, and a personal progress tracker showing which question types you're strongest in. Combine this guide with structured practice, and you'll confidently score 8–10/10 on Probability in your board exam.