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.
This theoretical (classical) probability is valid only when the outcomes are equally likely.
'A car starts or does not start' has two outcomes, but they are not equally likely, so the probability of starting is not ½. Always check that the listed outcomes have the same chance.
Experimental probability comes from what actually happened in trials. As the number of trials grows, it gets closer to the theoretical value.
The favourable outcomes can never be fewer than 0 or more than all outcomes, so
An event with exactly one outcome is an elementary event. The probabilities of all the elementary events of an experiment add up to 1.
Use the complement when 'not E' is easier to count, for example 'at least one six' = 1 − P(no six).
One coin: H, T (2 outcomes). Two coins: HH, HT, TH, TT (4). Three coins: 8 outcomes, best listed with a tree.
HT and TH are different outcomes. 'One head and one tail' therefore has probability 2/4 = ½, not ⅓.
One die: 1 to 6. Two dice (or one die thrown twice): 36 ordered pairs.
| Suit | Colour | Cards (13 each) |
|---|---|---|
| Spades ♠ | black | A, 2–10, J, Q, K |
| Clubs ♣ | black | A, 2–10, J, Q, K |
| Hearts ♥ | red | A, 2–10, J, Q, K |
| Diamonds ♦ | red | A, 2–10, J, Q, K |
For discs numbered 1 to n, count favourable numbers directly: multiples of k up to n number ⌊n/k⌋; perfect squares up to n number ⌊√n⌋.
If a point is equally likely to land anywhere in a region, the probability of landing in a part of it is the ratio of the areas.
| Idea | Result |
|---|---|
| Theoretical probability | favourable ÷ total (equally likely outcomes) |
| Range | 0 ≤ P(E) ≤ 1 |
| Sure / impossible | 1 / 0 |
| Complement | P(Ē) = 1 − P(E) |
| Sample spaces | coin 2, two coins 4, three coins 8, die 6, two dice 36, deck 52 |
Try each problem first, then open the solution to check your working.
A bag has 4 red, 6 blue and 5 green balls. One ball is drawn at random. Find the probability that it is (i) blue (ii) not green.
Given: 15 balls in all
Use: P(not E) = 1 − P(E)
P(blue) = 6/15 = 2/5
P(green) = 5/15 = 1/3, so P(not green) = 1 − 1/3 = 2/3
Answer: (i) 2/5 (ii) 2/3
A number is chosen at random from 1 to 20. Find the probability that it is a multiple of 2 or of 3.
Given: 20 equally likely numbers
Multiples of 2: 10 numbers. Multiples of 3 that are odd: 3, 9, 15 (3 numbers).
Favourable = 10 + 3 = 13
Answer: 13/20
The probability that a train is late is 0.36. What is the probability that it is not late?
Given: P(late) = 0.36
Use: P(Ē) = 1 − P(E)
1 − 0.36 = 0.64
Answer: 0.64
A bag holds 12 balls, each red or blue. Drawing a blue ball is twice as likely as drawing a red one. How many red balls are there?
Given: r red, 12 − r blue
P(blue) = 2 P(red) ⇒ (12 − r)/12 = 2r/12
12 − r = 2r ⇒ r = 4
Answer: 4 red balls
A box has 12 balls, of which x are red. If 6 more red balls are added, the probability of drawing a red ball doubles. Find x.
Given: before: x/12; after: (x + 6)/18
(x + 6)/18 = 2 · x/12 = x/6
x + 6 = 3x ⇒ x = 3
Answer: x = 3
What is the probability that a leap year, chosen at random, has 53 Sundays?
Given: 366 days = 52 weeks + 2 extra days
The 2 extra days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun): 7 equally likely pairs.
Two pairs contain a Sunday.
Answer: 2/7
Three coins are tossed together. Find the probability of (i) at least two heads (ii) exactly one head (iii) at least one head.
Given: 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
At least two heads: HHH, HHT, HTH, THH → 4
Exactly one head: HTT, THT, TTH → 3
At least one head = 1 − P(TTT) = 1 − 1/8
Answer: (i) 1/2 (ii) 3/8 (iii) 7/8
Two dice are thrown. Find the probability that the sum is (i) more than 9 (ii) 13.
Given: 36 outcomes
Sum 10: (4,6), (5,5), (6,4); sum 11: (5,6), (6,5); sum 12: (6,6) → 6 outcomes
A sum of 13 is impossible.
Answer: (i) 1/6 (ii) 0
Two dice are thrown. Find the probability that the product of the numbers is (i) 12 (ii) less than 10.
Given: 36 outcomes
Product 12: (2,6), (6,2), (3,4), (4,3) → 4 = 1/9
Product < 10: first die 1 → 6; 2 → 4 (2, 4, 6, 8); 3 → 3; 4 → 2; 5 → 1; 6 → 1; total 17
Answer: (i) 1/9 (ii) 17/36
Two dice are thrown. What is the probability that the numbers differ by 2?
Given: 36 outcomes
(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4) → 8
8/36 = 2/9
Answer: 2/9
Two dice are thrown. Find the probability that the sum is a prime number.
Given: possible prime sums: 2, 3, 5, 7, 11
Sum 2: 1 way; 3: 2; 5: 4; 7: 6; 11: 2
Favourable = 15, so P = 15/36
Answer: 5/12
Riya says that when two coins are tossed there are three outcomes (0, 1 or 2 heads), so each has probability 1/3. Is she right?
The outcomes 0, 1, 2 heads are not equally likely.
Sample space HH, HT, TH, TT: one head occurs twice (HT, TH).
Answer: No. P(1 head) = 1/2, P(0 heads) = P(2 heads) = 1/4
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) an ace (ii) a red card or a king (iii) a face card of clubs.
Given: 52 cards
Aces: 4 → 1/13
Red cards 26 plus the 2 black kings = 28 → 7/13
J, Q, K of clubs: 3
Answer: (i) 1/13 (ii) 7/13 (iii) 3/52
The jack, queen and king of spades are removed from a deck. One card is drawn from the rest. Find the probability that it is (i) a spade (ii) a face card (iii) a king.
Given: 49 cards remain
Spades left: 13 − 3 = 10
Face cards left: 12 − 3 = 9
Kings left: 4 − 1 = 3
Answer: (i) 10/49 (ii) 9/49 (iii) 3/49
A card is drawn from a deck of 52. What is the probability that it is a number card from 2 to 10?
Each suit has 9 such cards (2 to 10): 4 × 9 = 36
36/52
Answer: 9/13
Discs numbered 1 to 100 are in a box. One is drawn at random. Find the probability that its number is divisible by (i) 3 (ii) 3 or 5.
Given: 100 discs
Multiples of 3: 33
Multiples of 5: 20; of 15: 6. Divisible by 3 or 5: 33 + 20 − 6 = 47
Answer: (i) 33/100 (ii) 47/100
A number is chosen at random from 1 to 50. Find the probability that it is prime.
Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 → 15
15/50
Answer: 3/10
A dart lands at random on a square board of side 4 cm that has a circle of diameter 4 cm drawn inside it. Find the probability that it lands inside the circle. (Not for the board exam.)
Given: square area 16 cm², circle area π(2)² = 4π cm²
P = 4π/16 = π/4 ≈ 0.785
Answer: π/4
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Answer: (i) 1 (ii) 0, an impossible event (iii) 1, a sure (certain) event (iv) 1 (v) 0, 1
(i) Not equally likely: a car normally starts; it fails only when there is a fault.
(ii) Not equally likely: the result depends on the player's skill.
(iii) Equally likely: a true-false question has two answers, one right and one wrong.
(iv) Equally likely: a baby is (taken to be) equally likely to be a boy or a girl.
Answer: (iii) and (iv)
A fair coin has two equally likely outcomes, head and tail. The result cannot be predicted or influenced, so each team has the same chance, ½.
Answer: The outcomes are equally likely, so neither team is favoured
A probability lies between 0 and 1. 2/3, 15% = 0.15 and 0.7 all qualify; −1.5 is negative.
Answer: (B) −1.5
P(not E) = 1 − P(E) = 1 − 0.05
Answer: 0.95
(i) There is no orange candy: an impossible event.
(ii) Every candy is lemon: a sure event.
Answer: (i) 0 (ii) 1
'Same birthday' is the complement of 'not the same birthday'.
P = 1 − 0.992
Answer: 0.008
Total balls = 8.
(i) P(red) = 3/8
(ii) P(not red) = 1 − 3/8 = 5/8
Answer: (i) 3/8 (ii) 5/8
Total = 5 + 8 + 4 = 17.
(i) 5/17
(ii) 8/17
(iii) 1 − 4/17 = 13/17
Answer: (i) 5/17 (ii) 8/17 (iii) 13/17
Total coins = 100 + 50 + 20 + 10 = 180.
(i) P(50p) = 100/180 = 5/9
(ii) P(₹5) = 10/180 = 1/18, so P(not ₹5) = 1 − 1/18 = 17/18
Answer: (i) 5/9 (ii) 17/18
Total fish = 13, male = 5.
Answer: 5/13
8 equally likely outcomes.
(i) one outcome: 1/8
(ii) 1, 3, 5, 7: 4/8 = 1/2
(iii) 3, 4, 5, 6, 7, 8: 6/8 = 3/4
(iv) all 8 numbers: 8/8 = 1
Answer: (i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1
6 outcomes.
(i) Primes 2, 3, 5: 3/6 = 1/2
(ii) 3, 4, 5: 3/6 = 1/2
(iii) 1, 3, 5: 3/6 = 1/2
Answer: (i) 1/2 (ii) 1/2 (iii) 1/2
(i) King of hearts, king of diamonds: 2/52 = 1/26
(ii) 12 face cards: 12/52 = 3/13
(iii) 6 red face cards: 6/52 = 3/26
(iv) one card: 1/52
(v) 13 spades: 13/52 = 1/4
(vi) one card: 1/52
Answer: (i) 1/26 (ii) 3/13 (iii) 3/26 (iv) 1/52 (v) 1/4 (vi) 1/52
(i) One queen among 5 cards: 1/5
(ii) 4 cards remain: 10, J, K, A.
(a) P(ace) = 1/4
(b) No queen remains: P = 0
Answer: (i) 1/5 (ii) (a) 1/4 (b) 0
Total = 12 + 132 = 144.
P(good) = 132/144
Answer: 11/12
(i) 4/20 = 1/5
(ii) 19 bulbs remain, of which 4 are defective and 15 good: 15/19
Answer: (i) 1/5 (ii) 15/19
(i) Two-digit numbers 10 to 90: 81, so 81/90 = 9/10
(ii) 1, 4, 9, 16, 25, 36, 49, 64, 81: 9, so 9/90 = 1/10
(iii) 5, 10, …, 90: 18, so 18/90 = 1/5
Answer: (i) 9/10 (ii) 1/10 (iii) 1/5
6 faces. (i) A appears on 2 faces: 2/6 = 1/3
(ii) D appears on 1 face: 1/6
Answer: (i) 1/3 (ii) 1/6
Area of rectangle = 3 × 2 = 6 m². Area of circle = π(½)² = π/4 m².
P = (π/4) ÷ 6
Answer: π/24
Good pens = 144 − 20 = 124.
(i) 124/144 = 31/36
(ii) 20/144 = 5/36 (= 1 − 31/36)
Answer: (i) 31/36 (ii) 5/36
(i) Count the pairs giving each sum among the 36 outcomes:
| Sum on 2 dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
(ii) No. The 11 sums are not equally likely. For example, a sum of 2 comes only from (1, 1), while a sum of 7 comes from six pairs. So P(2) = 1/36 but P(7) = 6/36.
Answer: (ii) No; the sums are not equally likely
Outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (8).
He wins on HHH or TTT: P(win) = 2/8 = 1/4.
P(lose) = 1 − 1/4
Answer: 3/4
36 outcomes.
(i) Each throw has 5 non-5 results: 5 × 5 = 25 outcomes, so 25/36.
(ii) P(at least one 5) = 1 − 25/36 = 11/36. (The 11 outcomes are (5, 1) … (5, 6) and (1, 5) … (4, 5), (6, 5).)
Answer: (i) 25/36 (ii) 11/36
(i) Incorrect. The equally likely outcomes are HH, HT, TH, TT. 'One of each' covers HT and TH, so its probability is 2/4 = 1/2, while two heads and two tails each have 1/4.
(ii) Correct. Odd (1, 3, 5) and even (2, 4, 6) each contain 3 of the 6 equally likely faces, so P(odd) = 3/6 = 1/2.
Answer: (i) incorrect (ii) correct