Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Idea | Result |
|---|---|
| Range | 0 ≤ P(E) ≤ 1 |
| Impossible / certain | P = 0 / P = 1 |
| Experimental | times event occurred ÷ number of trials |
| Theoretical | favourable ÷ possible (equally likely) |
| Complement | P(not E) = 1 − P(E) |
| Two dice | 36 outcomes |
| Three coins | 8 outcomes |
| Independent trials | multiply along a tree path |
Try each problem first, then open the solution to check your working.
A fair die is rolled. Find the probability of a prime number.
Primes: 2, 3, 5 (3 outcomes) out of 6.
Answer: 1/2
A letter is chosen from MATHEMATICS. Find P(M) and P(vowel).
11 letters: M twice, vowels A, E, A, I (4).
Answer: P(M) = 2/11; P(vowel) = 4/11
A bag has 5 red, 3 blue and 2 green balls. Find the probability that a ball drawn is not blue.
10 balls; 7 are not blue.
Answer: 7/10
Cards numbered 1 to 20: find the probability of a multiple of 3.
Multiples: 3, 6, 9, 12, 15, 18 (6 of 20).
Answer: 3/10
Two dice are rolled. Find the probability that the sum is 8.
Pairs: (2,6), (3,5), (4,4), (5,3), (6,2): 5 of 36.
Answer: 5/36
A spinner has equal sectors numbered 1 to 8. Find the probability of a prime.
Primes: 2, 3, 5, 7 (4 of 8).
Answer: 1/2
Two dice are rolled. Give an event with probability 0 and an event with probability 1.
The sum is 1 (or 13) is impossible: P = 0.
The sum is between 2 and 12 is certain: P = 1.
Answer: 0 and 1
A die is rolled 60 times and 6 appears 14 times. Find the experimental probability of a 6 and compare with the theoretical probability.
Experimental = 14/60 = 7/30 ≈ 0.233.
Theoretical = 1/6 ≈ 0.167. The difference is normal for 60 trials; it should shrink as the trials increase.
Answer: 7/30 ≈ 0.233; 1/6 ≈ 0.167
In a survey of 80 students 20 prefer cricket. Estimate the number preferring cricket among 1200 students.
Relative frequency = 20/80 = 0.25.
0.25 × 1200 = 300.
Answer: about 300
A fair coin shows 5 heads in a row. What is the probability that the next toss is heads?
Each toss is independent; the coin has no memory.
Answer: 1/2
Three coins are tossed. Find the probability of at least two heads.
Outcomes (8): HHH, HHT, HTH, THH are favourable (4).
Answer: 1/2
A coin is tossed and a die is rolled. How many outcomes? Find P(head and an even number).
2 × 6 = 12 outcomes.
Favourable: H2, H4, H6 (3).
Answer: 12 outcomes; 1/4
A bag has 3 red and 2 blue balls. Two are drawn without replacement. Find P(both red).
P(first red) = 3/5; then 2 red among 4.
P = 3/5 × 2/4 = 3/10.
Answer: 3/10
A family has two children. List the sample space and find P(at least one girl).
S = {BB, BG, GB, GG}.
Favourable: BG, GB, GG.
Answer: 3/4
A dart lands at random on a 6 × 4 board. Find the probability that it lands in a 2 × 2 square marked on the board.
Area ratio = 4/24.
Answer: 1/6
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Because the outcome is random (cannot be predicted) and both sides are equally likely for a fair, unbiased coin. Neither team has an advantage.
Your friend can guess heads or tails but cannot know. All results are known beforehand but each individual toss is unpredictable. That is randomness.
Still 1/6 ≈ 0.167 for a fair die. Probability does not tell what happens next; it describes the long run.
To answer questions about the amount of rain we need more outcomes: {No Rain, Drizzle, Light Rain, Heavy Rain}. The sample space must be detailed enough for the question.
The outcomes HT and TH are favourable: 2 of 4. P = 1/2.
Certain (probability 1): Monday always follows Sunday.
Impossible (practically 0): Mumbai’s July weather is warm and rainy; snow has never been recorded.
Impossible, or at best extremely unlikely (very close to 0): classrooms are not accessible to elephants.
More likely (close to 1) if you attend school; you meet many friends every day. (It is not certain: you might be absent.)
(i) 8/30 = 4/15 ≈ 0.267. (ii) 7/30 × 600 = 140 yellow sweets.
(i) 11/40 = 0.275. (ii) 9/40 × 800 = 180 students.
Your results will vary. For example, 12 heads and 8 tails: (i) 12 (ii) 8 (iii) 12/20 = 0.6. (iv) The probability of tails on the next toss is 1/2, whatever happened before.
Perform the experiment and count. P(bottom) = (number landing on bottom)/100 and similarly for top and side. The three probabilities add up to 1. (Typical results show the side is the most likely.)
Favourable 2, 4, 6: 3/6 = 1/2.
(i) 3/12 = 1/4. (ii) 1/6. (iii) The sample of 12 is small, so chance variation is large. As the number of trials increases (60, 600, 6000) the experimental probability gets closer to 1/6 ≈ 0.167 (Law of Large Numbers).
6.
S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}: 12 outcomes.
Taking the ends as included: S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}: 11 outcomes. (If the ends were excluded: −4 to 4.)
S = {green, red} (colours). If we label all 12 balls: S = {G1, …, G5, R1, …, R7}, which has 12 equally likely outcomes. The two colours are not equally likely.
(i) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}: 6 outcomes. (ii) E = {(Samosa, Chai), (Samosa, Lassi)}.
First stage: Apple, Orange₁, Orange₂ from A. From each, second stage: Banana, Mango from B. 3 × 2 = 6 paths.
S = {(Apple, Banana), (Apple, Mango), (Orange₁, Banana), (Orange₁, Mango), (Orange₂, Banana), (Orange₂, Mango)}.
The six outcomes are equally likely (each fruit equally likely to be picked). One favourable: 1/6.
The first pen: R, B, G with probabilities 3/9, 4/9, 2/9; for each the second pen is again R, B, G. Colour pairs: RR, RB, RG, BR, BB, BG, GR, GB, GG (9 colour outcomes, not equally likely). With each of the 9 pens labelled there are 81 equally likely outcomes.
P = (3/9)² + (4/9)² + (2/9)² = (9 + 16 + 4)/81 = 29/81 ≈ 0.358.
(i) 0 (ii) sample space (iii) 1 (iv) 1/2.
The relative frequency is 15/50 = 0.3.
(i) No: the car is not equally likely to start or not. (ii) Yes: a fair coin. (iii) Yes: a fair die. (iv) No: 3 red and 7 blue marbles, so red and blue are not equally likely. (v) Boy or girl is usually treated as (nearly) equally likely, so we take it as approximately yes.
S = {HH, HT, TH, TT}; favourable 3: 3/4.
S = {1, …, 10}; even: 2, 4, 6, 8, 10: 5/10 = 1/2.
S = {1, …, 6}; favourable: 5, 6: 1/3.
6 balls; not red = 3: 1/2.
S has 8 outcomes; favourable HHT, HTH, THH: 3/8.
1/3.
| Jeans | Khakis | Shorts | |
|---|---|---|---|
| Red | Red + Jeans | Red + Khakis | Red + Shorts |
| Blue | Blue + Jeans | Blue + Khakis | Blue + Shorts |
2 × 3 = 6 outfits.
(i) 20/1000 = 0.02. (ii) (210 + 325)/1000 = 0.535. (iii) 445/1000 = 0.445.
(i) P, E, E, C = 4 of 5: 4/5. (ii) not E: P, A, C = 3 of 5: 3/5.
(i) 1/8 (ii) 4/8 = 1/2 (iii) 6/8 = 3/4 (iv) all 8: 1 (v) 3 and 6: 2/8 = 1/4.
First: R 4/9, B 5/9. After R: 3 red, 5 blue: R 3/8, B 5/8. After B: 4 red, 4 blue: R 4/8, B 4/8. (i) 4/9 × 5/8 = 20/72 = 5/18. (ii) 5/9 × 4/8 = 20/72 = 5/18.
Event of probability 0: “the sum is 13” (or “the sum is 1”). For probability 1 we need a certain event: “the sum is between 2 and 12”. No single outcome has probability 1, since each of the 36 pairs has probability 1/36.
Primes greater than 5 up to 12: 7 and 11. Sum 7: 6 ways; sum 11: 2 ways. P = 8/36 = 2/9.
Same colour: RR 6, GG 3, BB 1 = 10 of C(9,2) = 36 pairs. Different = 26/36 = 13/18.
Favourable: HHT, HTH. P = 2/8 = 1/4.
24 numbers. Even: last digit 2 or 4: 2 × 3! = 12. P = 12/24 = 1/2.
4³ = 64 equally likely answer patterns. Exactly 2 right: choose the position of the wrong one (3 ways) and its wrong option (3 ways): 9. P = 9/64.
(i) With replacement: 4 first-draw branches, each with 4 second-draw branches: 16 outcomes (1,1), (1,2), …, (4,4). (ii) Without replacement: 4 first branches, each with 3 second branches: 12 outcomes (no pairs like (1,1)). (iii) The sizes are 16 and 12.
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}: 12 elements.
(iv) {0, 1, 2, 3} is correct. (i) {1, 2, 3} omits 0 (all tails). (ii) {0, 1, 2} omits 3 (all heads). (iii) {0, 1, 2, 3, 4} contains 4, which cannot happen.
Circle area = π(0.5)² = π/4. Rectangle area = 6. P = (π/4)/6 = π/24 ≈ 0.131.