NCERT Solutions Class 9 Maths Chapter 7: Probability

(i) Certain, 1. (ii) Impossible, 0. (iii) Impossible (practically 0). (iv) More likely, about 0.9.

Answers for (iv) can differ a little (0.8 to 0.95): any number clearly above 0.5 and below 1 with a good reason is acceptable.

P(4) = 16 ≈ 0.167.

1. Favourable outcomesOnly one face shows 4, so 1.
2. Total outcomesThe die has 6 faces: 1, 2, 3, 4, 5, 6.
3. ProbabilityP(4) = 16 ≈ 0.167 (about 16.7%)

P(B) = 211 ≈ 0.182.

1. Total lettersP-R-O-B-A-B-I-L-I-T-Y has 11 letters.
2. Letter BB appears 2 times.
3. ProbabilityP(B) = 211 ≈ 0.182

(i) P(green) = 415. (ii) About 140 yellow sweets.

1. (i) Green sweetsP(green) = 830 = 415
2. (ii) Yellow sweetsFrom the sample, P(yellow) = 730.Estimate = 730 × 600 = 7 × 20 = 140

(i) P(Arts) = 1140 = 0.275. (ii) About 180 students prefer the Sports Club.

1. (i) Arts ClubP(Arts) = 1140 = 0.275 (27.5%)
2. (ii) Sports Club in the whole schoolIn the sample, the fraction who like Sports is 940.940 × 800 = 9 × 20 = 180

The sample is used to estimate the whole school, without asking all 800 students.

Your numbers will depend on your own tosses. For example, 11 heads and 9 tails give P(heads) = 0.55, and the next toss still has P(tails) = 1/2.

1. What to doToss a coin 20 times and tally the results in a table.
2. Example data (yours will differ)

   Result	Tally	Count
   Heads	||||| ||||| |	11
   Tails	||||| ||||	9

3. (iii) Experimental probabilityP(heads) = number of heads20 = 1120 = 0.55(use your own count of heads).
4. (iv) The next tossEach toss is independent, so past results do not change the next one: P(tails) = 1/2 = 0.5.

This is an activity question, so the answers (i) to (iii) are your own results. The numbers 11 and 9 above are only an example.

Experimental probability = (number of times an outcome occurs) ÷ 100. Your counts will be your own; the table shows an example.

1. What to doToss the cup 100 times and record where it lands each time. Count each outcome.
2. Example data (yours will differ)

   Outcome	Frequency	Experimental probability
   On its side	70	70100 = 0.7
   On its bottom	20	20100 = 0.2
   Upside down (on its top)	10	10100 = 0.1
   Total	100	1

3. CheckThe three probabilities add up to 1.

Unlike a fair coin, a paper cup is not equally likely to land every way because of its shape. So we cannot find these probabilities by theory; we must measure them by experiment.

The numbers 70, 20 and 10 are only an example. Your own experiment will give different counts.

P(even) = 12.

1. Sample spaceS = {1, 2, 3, 4, 5, 6}, so 6 outcomes.
2. Favourable outcomesThe even numbers are 2, 4, 6: 3 outcomes.
3. ProbabilityP(even) = 36 = 12 = 0.5 (50%)

(i) 3/12 = 0.25. (ii) 1/6 ≈ 0.167. With more rolls, the experimental value moves closer to 1/6.

1. (i) ExperimentalP(3) = 312 = 14 = 0.25
2. (ii) TheoreticalP(3) = 16 ≈ 0.167
3. (iii) Why they differWith only 12 rolls, chance can easily push the result away from the expected value. As the number of rolls grows to 60, 600 or 6000, the experimental probability gets closer and closer to 1/6. This is the law of large numbers.

There are 3 × 2 = 6 combinations. The event “Samosa” has 2 outcomes.

1. (i) Sample spaceEach snack can go with either drink, so there are 3 × 2 = 6 pairs.S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}
2. (ii) The event “Samosa”E = {(Samosa, Chai), (Samosa, Lassi)}

There are 6 possible outcomes: S = {1, 2, 3, 4, 5, 6}.

The die has six faces numbered 1 to 6, and no other result is possible. So the number of elements in the sample space is n(S) = 6.

(i) 12 outcomes. (ii) 9 integers. (iii) {Green, Red}, or the 12 balls.

1. (i) A die and a coinEach of the 6 numbers pairs with H or T: S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}, 12 outcomes.
2. (ii) Integers between −5 and +5S = {−4, −3, −2, −1, 0, 1, 2, 3, 4}, 9 outcomes (−5 and +5 are not included).
3. (iii) Balls in a boxBy colour: S = {Green, Red}. These two outcomes are not equally likely: P(Green) = 5/12 and P(Red) = 7/12. If we label all 12 balls (G₁ to G₅, R₁ to R₇), the 12 outcomes are equally likely.

Yes. When two coins are tossed, P(one head and one tail) = 1/2.

1. Sample space (tree diagram)The outcomes are HH, HT, TH and TT: 4 equally likely outcomes.
2. Favourable outcomesHT and TH: 2 outcomes.
3. ProbabilityP = 24 = 12 = 0.5

The question does not say how many coins; we take a coin tossed twice (or two coins tossed together).

The sample space has 6 pairs, and P(apple and banana) = 1/6.

1. (i) Tree diagramTreat the two oranges as different (O₁, O₂) so the 3 fruits in basket A are equally likely, each with probability 1/3. From each, the fruit in basket B is Banana or Mango, each with probability 1/2.
2. (ii) Sample spaceS = {(A, B), (A, M), (O₁, B), (O₁, M), (O₂, B), (O₂, M)},   n(S) = 6
3. (iii) Apple and bananaOnly (A, B) matches.P = 13 × 12 = 16 ≈ 0.167

There are 9 colour pairs, and P(same colour) = 29/81.

1. (i) Possible outcomesThere are 3 + 4 + 2 = 9 pens. Because the pen is put back, both picks have the same chances: P(R) = 3/9, P(B) = 4/9, P(G) = 2/9. The 9 outcomes are (R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G).
2. (ii) Same colourP(R,R) = 39 × 39 = 981P(B,B) = 49 × 49 = 1681P(G,G) = 29 × 29 = 481P(same colour) = 9 + 16 + 481 = 2981 ≈ 0.358

0

An impossible event never happens, so its probability is 0.

sample space

The set of all possible outcomes of a random experiment is called the sample space.

1

A certain event always happens, so its probability is 1.

0.5 (that is, 1/2)

There are 2 equally likely outcomes, and 1 of them is heads: P(heads) = 1/2 = 0.5.

relative frequency, 15/50 = 0.3

The number 15 is the frequency. The relative frequency is frequency ÷ total = 1550 = 310 = 0.3.

(ii), (iii) and (v, approximately) have equally likely outcomes. (i) and (iv) do not.

(i) 3/4 (ii) 1/2 (iii) 1/3 (iv) 1/2 (v) 3/8

1. (i) Two coinsS = {HH, HT, TH, TT}. “At least one head”: HH, HT, TH.P = 34 = 0.75
2. (ii) Cards 1 to 10S = {1, 2, …, 10}. Even: 2, 4, 6, 8, 10.P = 510 = 12
3. (iii) Greater than 4S = {1, …, 6}. Favourable: 5, 6.P = 26 = 13 ≈ 0.33
4. (iv) Not redThere are 3 + 2 + 1 = 6 balls; not red means blue or green: 2 + 1 = 3.P = 36 = 12
5. (v) Three coinsS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, 8 outcomes. Exactly two heads: HHT, HTH, THH.P = 38 = 0.375

P(strawberry) = 1/3.

The sample space has 3 equally likely outcomes, and 1 is strawberry: P = 13 ≈ 33.3%.

There are 2 × 3 = 6 possible outfits.

(i) 0.02 (ii) 0.535 (iii) 0.445

1. Total cases20 + 210 + 325 + 445 = 1000.
2. (i) Less than 4000 kmP = 201000 = 0.02
3. (ii) Between 4000 and 14000 kmAdd the two middle groups: 210 + 325 = 535.P = 5351000 = 0.535
4. (iii) More than 14000 kmP = 4451000 = 0.445

Check: 0.02 + 0.535 + 0.445 = 1.

(i) 4/5 = 0.8 (ii) 3/5 = 0.6

1. The cardsP, E, A, C, E: 5 cards (E appears twice).
2. (i) P, E or COne P, two E’s and one C: 4 cards.P = 45 = 0.8
3. (ii) Not an EThe cards P, A, C: 3 cards.P = 35 = 0.6

(i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1 (v) 1/4

(i) 5/18 (ii) 5/18

1. Why the second draw changesThe first ball is not put back, so the second draw is from 8 balls. If the first was red, 3 red and 5 blue remain; if it was blue, 4 red and 4 blue remain.
2. (i) Red then blueP = 49 × 58 = 2072 = 518
3. (ii) Two blueP = 59 × 48 = 2072 = 518

Probability 0: the sum is 13. Probability 1: the sum is between 2 and 12.

1. An event with probability 0The smallest sum is 1 + 1 = 2 and the largest is 6 + 6 = 12, so “the sum is 13” can never happen: P = 0.
2. An event with probability 1“The sum is a number from 2 to 12” (or “the sum is at least 2”) always happens: P = 1.

(i) 2/9 (ii) 13/18 (iii) 1/4 (iv) 1/2 (v) 9/64

1. (i) Prime sum greater than 5Two dice give 6 × 6 = 36 outcomes. Primes greater than 5 are 7 and 11.Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1): 6 outcomesSum 11: (5,6), (6,5): 2 outcomesP = 836 = 29 ≈ 0.22
2. (ii) Different coloursThere are 4 + 3 + 2 = 9 balls. It is easier to find the probability of the same colour and subtract from 1.P(RR) = 49 × 38 = 1272,   P(GG) = 39 × 28 = 672,   P(BB) = 29 × 18 = 272P(same) = 2072,   P(different) = 1 − 2072 = 5272 = 1318 ≈ 0.72
3. (iii) First head and exactly two headsS = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Outcomes that start with H and have exactly two heads: HHT, HTH.P = 28 = 14
4. (iv) Even four-digit numberThere are 4! = 24 numbers. An even number must end in 2 or 4. Ending in 2: 3! = 6 numbers; ending in 4: 6 numbers. So 12 even numbers.P = 1224 = 12
5. (v) Exactly 2 correct guessesFor each question P(correct) = 1/4 and P(wrong) = 3/4. Two right and one wrong can happen in 3 ways (CCW, CWC, WCC).P = 3 × 14 × 14 × 34 = 964 ≈ 0.14

(i) 16 outcomes (with replacement). (ii) 12 outcomes (without replacement).

1. (i) With replacementEach of the 4 first branches (1, 2, 3, 4) splits into 4 branches, because the first ball is put back.

   First	Second draw gives
   1	(1,1) (1,2) (1,3) (1,4)
   2	(2,1) (2,2) (2,3) (2,4)
   3	(3,1) (3,2) (3,3) (3,4)
   4	(4,1) (4,2) (4,3) (4,4)

2. (ii) Without replacementEach first branch splits into only 3 branches, because the first ball cannot be drawn again.

   First	Second draw gives
   1	(1,2) (1,3) (1,4)
   2	(2,1) (2,3) (2,4)
   3	(3,1) (3,2) (3,4)
   4	(4,1) (4,2) (4,3)

3. (iii) SizesWith replacement: 4 × 4 = 16   and   without replacement: 4 × 3 = 12

Only (iv) {0, 1, 2, 3} is a sample space.

P = π ÷ 24 ≈ 0.131.

1. Area of the rectangle (all outcomes)3 m × 2 m = 6 m²
2. Area of the circle (favourable outcomes)Diameter 1 m, so radius 0.5 m.π × (0.5)² = 0.25π m²
3. Probability = area of the circle ÷ area of the rectangle0.25π6 = π24 ≈ 0.131

The dye can land anywhere in the rectangle, so the chance of landing in a region is the ratio of the areas.