This MCQ module is based on: Probability — Exercises
TOPIC 22 OF 25
Probability — Exercises
🎓 Class 9
Mathematics
CBSE
Theory
Ch 7 — The Mathematics of Maybe: Introduction to Probability
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Probability — Exercises
Targeting Class 9 level in Probability, with Intermediate difficulty.
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Chapter 7 Summary — Key Ideas at a Glance
Quick recap
- Probability measures the chance of an event, with values in [0, 1].
- Sample space (S) = set of all possible outcomes of an experiment.
- Classical: \(P(E) = n(E)/n(S)\) when outcomes are equally likely.
- Empirical: \(P(E) = m/n\) from observed trials.
- Subjective: probability from belief or expert judgement.
- Complement rule: \(P(E) + P(\bar E) = 1\).
- Certain event has P = 1; impossible event has P = 0.
Exercise 7.1 — Sample Space and Classical Probability
Q1. Write the sample space for each experiment: (i) tossing a coin twice, (ii) rolling two distinguishable dice, (iii) drawing one ball from a bag containing balls numbered 1–5.
(i) S = {HH, HT, TH, TT}, |S| = 4. (ii) S = {(a,b) : 1 ≤ a, b ≤ 6}, |S| = 36. (iii) S = {1, 2, 3, 4, 5}, |S| = 5.
Q2. A fair die is rolled once. Find the probability of getting (i) a multiple of 3, (ii) a number greater than 4, (iii) an odd prime number.
(i) Multiples of 3 in {1..6} are {3, 6}: P = 2/6 = 1/3. (ii) Greater than 4: {5, 6}: P = 2/6 = 1/3. (iii) Odd primes ≤ 6: {3, 5}: P = 2/6 = 1/3.
Q3. A bag contains 4 red, 5 white and 6 black balls. A ball is drawn at random. Find the probability that it is (i) white, (ii) not white, (iii) black or red.
Total = 15. (i) White: 5/15 = 1/3. (ii) Not white: 10/15 = 2/3. (iii) Black or red: 6 + 4 = 10. P = 10/15 = 2/3.
Q4. One card is drawn at random from a deck of 52. Find the probability of getting (i) a spade, (ii) a red king, (iii) a card with a number 2–10 (i.e., neither face card nor ace).
(i) Spades = 13. P = 13/52 = 1/4. (ii) Red kings = 2 (heart, diamond). P = 2/52 = 1/26. (iii) Numbers 2–10 in 4 suits = 9 × 4 = 36. P = 36/52 = 9/13.
Q5. Two coins are tossed simultaneously. Find the probability of getting (i) exactly one head, (ii) at least one head, (iii) no heads.
S = {HH, HT, TH, TT}, |S| = 4. (i) Exactly one H: {HT, TH}, P = 2/4 = 1/2. (ii) At least one H: {HH, HT, TH}, P = 3/4. (iii) No heads: {TT}, P = 1/4.
Q6. A letter is chosen at random from the word "MATHEMATICS". Find the probability that it is (i) a vowel, (ii) the letter M, (iii) a consonant.
Letters: M, A, T, H, E, M, A, T, I, C, S → 11 letters. Vowels = A, E, A, I = 4. Ms = 2. Consonants = 11 - 4 = 7. (i) 4/11. (ii) 2/11. (iii) 7/11.
Exercise 7.2 — Empirical Probability
Q7. A coin was tossed 1000 times. Heads appeared 524 times. (i) Find the empirical probability of head. (ii) Compare with the classical probability and comment.
(i) P(H) = 524/1000 = 0.524. (ii) Classical P(H) = 0.5. The empirical value is very close to 0.5, suggesting the coin is fair. The slight difference is due to natural sampling fluctuation.
Q8. The blood groups of 60 students of a class were recorded:
A student is chosen at random. Find the probability that the student has blood group (i) A, (ii) AB, (iii) not O.
| Group | A | B | AB | O |
|---|---|---|---|---|
| Students | 21 | 18 | 6 | 15 |
n = 60. (i) P(A) = 21/60 = 7/20. (ii) P(AB) = 6/60 = 1/10. (iii) P(not O) = (60 - 15)/60 = 45/60 = 3/4.
Q9. In a survey of 200 households, 90 read newspaper P, 60 read newspaper Q (only), and 50 read neither. (i) What fraction of households read P? (ii) What is the probability that a randomly chosen household reads at least one newspaper?
(i) 90/200 = 9/20. (ii) Read at least one = 200 - 50 = 150. P = 150/200 = 3/4.
Q10. Out of 250 patients tested for a disease, 32 tested positive. Estimate the empirical probability that the next patient tested also turns out positive. State one important assumption.
P ≈ 32/250 = 0.128 = 12.8%. Assumption: the next patient comes from the same population (same risk factors) as the 250 already tested.
Exercise 7.3 — Mixed Higher-Order Problems
Q11. Two dice are rolled. Find the probability of: (i) doublet, (ii) sum = 8, (iii) at least one 6.
|S| = 36. (i) Doublets = (1,1)…(6,6) = 6. P = 6/36 = 1/6. (ii) Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5. P = 5/36. (iii) At least one 6: complement of "no 6" = 1 - 25/36 = 11/36.
Q12. A box contains 12 bulbs, 3 of which are defective. One bulb is taken at random. Find the probability that it is (i) defective, (ii) good. If the first bulb taken is good and not replaced, find the probability that the next bulb is also good.
(i) P(def) = 3/12 = 1/4. (ii) P(good) = 9/12 = 3/4. After removing one good bulb: 11 left, 8 good. P(next good) = 8/11.
Q13. A spinner has sectors 1–10 of equal size. Find the probability of (i) prime number, (ii) number divisible by 4, (iii) two-digit number.
|S| = 10. (i) Primes: {2,3,5,7} = 4. P = 4/10 = 2/5. (ii) Divisible by 4: {4, 8} = 2. P = 1/5. (iii) Two-digit: {10} = 1. P = 1/10.
Q14. A meteorologist says, "There is an 80% chance of rain tomorrow." (i) Is this classical, empirical or subjective? (ii) What is the probability that it will not rain?
(i) Subjective — based on expert judgement using a model and past data. (ii) P(no rain) = 1 - 0.80 = 0.20 = 20%.
Q15. From a deck of 52, two cards are drawn one after the other without replacement. (i) What is the probability the first is an ace? (ii) Given that the first is an ace, find the probability that the second is also an ace.
(i) 4/52 = 1/13. (ii) After removing one ace, 51 cards remain with 3 aces. P = 3/51 = 1/17.
Activity 7.3 — A Mini Probability Project
L5 EvaluateMaterials: Sticky notes, dice, calculator, pen, classmates.
Goal: Use both classical and empirical probability for the same experiment and compare.
- Pick an experiment, e.g., "rolling two dice and recording the sum."
- Compute the classical probability of each sum from 2 to 12. (Use Fig. 7.6 logic.)
- Roll the two dice 50 times and record each sum on a tally sheet.
- Compute the empirical probability of each sum.
- Plot both sets of probabilities on the same bar chart. Discuss the differences and what would happen at 500 trials.
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Competency-Based Questions
Scenario: A factory makes torch bulbs. A quality inspector tested 500 bulbs and found that 23 were defective. The factory ships bulbs in cartons of 100 each.
Q1. Estimate the empirical probability that a random bulb from this factory is defective. Roughly, how many defective bulbs would you expect in a carton of 100?
L3 ApplyAnswer: P(defective) = 23/500 = 0.046. Expected defectives in a carton of 100 ≈ 0.046 × 100 ≈ 4 to 5 bulbs.
Q2. The factory advertises "More than 95% of our bulbs work perfectly." Analyse whether the data supports this claim.
L4 AnalyseAnswer: P(working) = 1 - 0.046 = 0.954 = 95.4%. The data just barely supports the claim. A larger sample would confirm it; the value is close enough to 95% that fluctuation matters.
Q3. A retailer claims, "If I buy 100 bulbs, exactly 5 will be defective." Evaluate whether this is a sound conclusion from the empirical probability.
L5 EvaluateAnswer: No. The probability gives an expected average, not a guarantee. In any one carton you may get 0, 3, 7 or even 12 defectives. Probability describes long-run frequency, not exact counts in a single sample.
Q4. Design a quality-improvement plan: what new data should the factory collect, and how should it use empirical probability to track and reduce the defect rate over time?
L6 CreateOne plan: (1) Test a sample of 200 bulbs every week from each production line. (2) Compute weekly P(defective). (3) Plot the time series; flag any week where P > 0.05 for investigation. (4) Record causes (filament, glass, wiring) for every defective bulb. (5) After fixes, compare the new month's P to the previous month — P should fall.
Assertion–Reason Questions
Assertion (A): P(rolling a 7 on a standard die) = 0.
Reason (R): 7 is not part of the sample space {1, 2, 3, 4, 5, 6}.
Reason (R): 7 is not part of the sample space {1, 2, 3, 4, 5, 6}.
Answer: (a) — Both true; R explains A directly.
Assertion (A): If P(E) = 0.4, then P(not E) = 0.6.
Reason (R): P(E) and P(not E) are always equal.
Reason (R): P(E) and P(not E) are always equal.
Answer: (c) — A is true (1 - 0.4 = 0.6). R is false: P(E) and P(not E) sum to 1 but are not equal in general.
Assertion (A): Empirical probabilities computed from 10 trials are very reliable.
Reason (R): By the law of large numbers, empirical probability approaches theoretical probability as trials increase.
Reason (R): By the law of large numbers, empirical probability approaches theoretical probability as trials increase.
Answer: (d) — A is false (10 trials are too few). R is a true statement which actually contradicts A — reliability requires large n.
Glossary — Key Terms
Quick definitions
Random experiment: action with unpredictable result.
Outcome: single result of a trial.
Sample space (S): set of all possible outcomes.
Event (E): subset of the sample space.
Equally likely: outcomes with equal probability.
Classical probability: \(n(E)/n(S)\) under fairness.
Empirical probability: ratio of favourable trials to total trials in real data.
Subjective probability: probability assigned by judgement.
Complement: "not E"; \(P(\bar E) = 1 - P(E)\).
Frequently Asked Questions
What is the probability of getting a king when drawing a card from a 52-card deck?
There are 4 kings in a 52-card deck, so P(king) = 4/52 = 1/13 ≈ 0.0769.
What is the probability of getting a sum of 7 when rolling two dice?
Total outcomes = 6 x 6 = 36. Favourable outcomes for sum 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes. P(sum = 7) = 6/36 = 1/6.
If a bag contains 3 red and 5 blue balls, what is P(red)?
Total balls = 3 + 5 = 8. Favourable outcomes (red) = 3. So P(red) = 3/8 = 0.375.
What is the probability of getting at least one head when tossing two fair coins?
Sample space S = {HH, HT, TH, TT}, four equally likely outcomes. 'At least one head' includes HH, HT, TH — three outcomes. P(at least one head) = 3/4 = 0.75. You can also use P(at least one head) = 1 - P(no heads) = 1 - 1/4 = 3/4.
If P(E) = 0.35, what is P(not E)?
By the complementary event rule, P(not E) = 1 - P(E) = 1 - 0.35 = 0.65.
In 200 traffic-light observations, the light was red 80 times. What is the empirical probability that the light is red?
Empirical probability = (red observations)/(total observations) = 80/200 = 0.4 = 40%.
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