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NCERT Class 9 Mathematics The Mathematics of Maybe: Introduction to Probability: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “The Mathematics of Maybe: Introduction to Probability”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Fill in the blanks: (i) probability of an impossible event; (ii) the set of all outcomes of an experiment; (iii) probability of a certain event; (iv) probability of heads with a fair coin.

   1. (i) An impossible event has no outcomes, so its probability is 0.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   2. (ii) The collection of all possible outcomes of an experiment is called the sample space.
   3. (iii) A certain event is guaranteed to happen, so its probability is 1.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   4. (iv) A fair coin has two equally likely outcomes (heads and tails), so the probability of heads is 1/2.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   Answer: (i) 0; (ii) sample space; (iii) 1; (iv) 1/2

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2. Exercises Q2

   15 of 50 surveyed students like football. 15 is the frequency; name the other measure (frequency or relative frequency) and give its value.

   1. The total number of students surveyed is 50.
   2. The number of students who like football is 15.
   3. The measure that is not the frequency is the relative frequency.
   4. Relative frequency is calculated by dividing the frequency by the total number of observations.
   5. Relative frequency = 15 ÷ 50.
   6. Simplifying the fraction gives 15/50.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Converting the fraction to a decimal gives 0.3.

   Answer: relative frequency; 15/50; 0.3

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3. Exercises Q3

   Which experiments have equally likely outcomes, and why? (i) a car starts or does not start; (ii) tossing a fair coin; (iii) rolling a fair die; (iv) a marble from a bag of 3 red and 7 blue; (v) a baby is a boy or a girl.

   1. (i) A car starting or not depends on its condition, so the chances are not equal.
   2. (ii) A fair coin has two sides, heads and tails, which are equally likely to appear.
   3. (iii) A fair die has six faces numbered 1 to 6, and each face has an equal chance of coming up.
   4. (iv) The bag has 3 red and 7 blue marbles, so the chance of drawing a blue marble is higher than a red one.
   5. (v) Boys and girls are born in nearly equal numbers, but not exactly: slightly more boys are born (about 105 boys for every 100 girls worldwide), so the two outcomes are close to, but not exactly, equally likely.

   Answer: (i) No; (ii) Yes; (iii) Yes; (iv) No (blue is more likely); (v) Nearly equally likely, but not exactly

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4. Exercises Q4

   Sample space and probability: (i) two coins, at least one head; (ii) cards 1 to 10, an even number; (iii) one die, more than 4; (iv) bag of 3 red, 2 blue, 1 green, not red; (v) three coins, exactly two heads.

   1. (i) For two coins, the sample space is {HH, HT, TH, TT} with 4 outcomes.
   2. (i) The outcomes with at least one head are {HH, HT, TH}, so there are 3 favourable outcomes.
   3. (i) Probability = 3/4.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   4. (ii) For cards 1 to 10, the total outcomes are 10.
   5. (ii) The even numbers are {2, 4, 6, 8, 10}, so there are 5 favourable outcomes.
   6. (ii) Probability = 5/10 = 1/2.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   7. (iii) For one die, the sample space is {1, 2, 3, 4, 5, 6} with 6 outcomes.
   8. (iii) The outcomes more than 4 are {5, 6}, so there are 2 favourable outcomes.
   9. (iii) Probability = 2/6 = 1/3.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   10. (iv) The bag has 3 red, 2 blue, and 1 green ball, making a total of 6 balls.
   11. (iv) The outcomes not red are the 2 blue and 1 green balls, so there are 3 favourable outcomes.
   12. (iv) Probability = 3/6 = 1/2.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   13. (v) For three coins, the sample space has 2³ = 8 outcomes.
   14. (v) The outcomes with exactly two heads are {HHT, HTH, THH}, so there are 3 favourable outcomes.
   15. (v) Probability = 3/8.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

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

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5. Exercises Q5

   A bag has 3 candies (strawberry, lemon, mint). Probability of picking the strawberry one.

   1. Total number of candies in the bag = 3.
   2. Number of strawberry candies = 1.
   3. Probability of picking a strawberry candy = (Number of strawberry candies) / (Total number of candies).
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   4. Probability = 1/3.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   Answer: 1/3

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6. Exercises Q6

   2 shirts (red, blue) and 3 pants (jeans, khakis, shorts): list every one-shirt-one-pant outfit in a table.

   1. Pair each of the 2 shirts with each of the 3 pants: 2 × 3 = 6 outfits.
   2. Table, shirt by pants: Red shirt | Red + Jeans | Red + Khakis | Red + Shorts.
   3. Blue shirt | Blue + Jeans | Blue + Khakis | Blue + Shorts.

   Answer: 6 outfits: Red + Jeans, Red + Khakis, Red + Shorts, Blue + Jeans, Blue + Khakis, Blue + Shorts

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7. Exercises Q7

   1000 tyres by distance before replacement: under 4000 km 20; 4001 to 9000 km 210; 9001 to 14000 km 325; over 14000 km 445. Probability a tyre lasts (i) under 4000 km, (ii) between 4000 and 14000 km, (iii) over 14000 km.

   1. Total number of tyres tested is 1000.
   2. (i) Number of tyres lasting under 4000 km is 20.
   3. (i) Probability = 20 ÷ 1000 = 0.02.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   4. (ii) Number of tyres lasting between 4000 and 14000 km is 210 + 325 = 535.
   5. (ii) Probability = 535 ÷ 1000 = 0.535.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   6. (iii) Number of tyres lasting over 14000 km is 445.
   7. (iii) Probability = 445 ÷ 1000 = 0.445.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   Answer: (i) 0.02; (ii) 0.535; (iii) 0.445

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8. Exercises Q8

   The letters of PEACE are on five cards; one is drawn. Probability that it is (i) P, E or C; (ii) not an E.

   1. The word PEACE has 5 letters: P, E, A, C, E.
   2. Total number of possible outcomes when one card is drawn is 5.
   3. (i) The letters P, E, or C appear in the word as P, E, A, C, E.
   4. (i) The distinct letters among P, E, or C are P, E, and C.
   5. (i) The number of cards showing P, E, or C is 4 (one P, two Es, one C).
   6. (i) Probability of drawing P, E, or C is 4/5.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   7. (ii) The letters that are not E are P, A, and C.
   8. (ii) The number of cards showing a letter other than E is 3.
   9. (ii) Probability of drawing a letter that is not E is 3/5.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   Answer: (i) 4/5; (ii) 3/5

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9. Exercises Q9

   A spinner stops on 1 to 8, all equally likely (Fig. 7.7). Probability of (i) 8, (ii) an odd number, (iii) more than 2, (iv) less than 9, (v) a multiple of 3.

   1. The spinner has 8 equally likely outcomes: 1, 2, 3, 4, 5, 6, 7, 8.
   2. (i) The number of favourable outcomes for getting 8 is 1.
   3. (i) Probability of getting 8 is 1/8.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   4. (ii) The odd numbers are 1, 3, 5, 7, so there are 4 favourable outcomes.
   5. (ii) Probability of getting an odd number is 4/8, which simplifies to 1/2.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   6. (iii) The numbers more than 2 are 3, 4, 5, 6, 7, 8, so there are 6 favourable outcomes.
   7. (iii) Probability of getting a number more than 2 is 6/8, which simplifies to 3/4.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   8. (iv) All numbers from 1 to 8 are less than 9, so there are 8 favourable outcomes.
   9. (iv) Probability of getting a number less than 9 is 8/8, which simplifies to 1.
      why?

      Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

   10. (v) The multiples of 3 are 3 and 6, so there are 2 favourable outcomes.
   11. (v) Probability of getting a multiple of 3 is 2/8, which simplifies to 1/4.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

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

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10. Exercises Q10

    A basket has 4 red and 5 blue balls; one ball is drawn and kept aside, then a second is drawn. Tree diagram, then (i) P(red then blue), (ii) P(two blue).

    1. Total balls in the basket = 4 red + 5 blue = 9.
    2. Probability of drawing a red ball first = 4/9.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    3. Probability of drawing a blue ball first = 5/9.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    4. If a red ball is drawn first, 8 balls remain with 5 blue balls.
    5. Probability of drawing a blue ball second given red first = 5/8.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    6. (i) P(red then blue) = P(red first) × P(blue second | red first) = (4/9) × (5/8).
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    7. (i) P(red then blue) = 20/72 = 5/18.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    8. If a blue ball is drawn first, 8 balls remain with 4 blue balls.
    9. Probability of drawing a blue ball second given blue first = 4/8.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    10. (ii) P(two blue) = P(blue first) × P(blue second | blue first) = (5/9) × (4/8).
        why?

        Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    11. (ii) P(two blue) = 20/72 = 5/18.
        why?

        Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    Answer: (i) 5/18; (ii) 5/18

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11. Exercises Q11

    A pair of six-sided dice is thrown: give an event with probability 0 and one with probability 1.

    1. A pair of six-sided dice has 36 possible outcomes when thrown.
    2. An event with probability 0 is an impossible event, meaning it cannot happen.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    3. An example of an impossible event is getting a sum of 13, as the maximum sum is 12.
    4. An event with probability 1 is a sure event, meaning it is certain to happen.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    5. An example of a sure event is getting a sum between 2 and 12, as all outcomes fall in this range.

    Answer: Event with probability 0: Getting a sum of 13; Event with probability 1: Getting a sum between 2 and 12

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12. Exercises Q12

    (i) Two dice: P(sum is a prime greater than 5). (ii) Bag of 4 red, 3 green, 2 blue; two balls drawn without replacement: P(different colours). (iii) Three coins: P(first is heads and exactly two heads). (iv) Four-digit numbers from 1, 2, 3, 4 without repetition: P(even). (v) 3 questions with 4 options each, all guessed: P(exactly 2 correct).

    1. (i) Total outcomes for two dice are 36.
    2. (i) Prime sums greater than 5 are 7 and 11.
    3. (i) Outcomes for 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) and for 11 are (5,6), (6,5), totaling 8.
    4. (i) Probability is 8/36 which simplifies to 2/9.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    5. (ii) Total ways to draw 2 balls from 9 are 36.
    6. (ii) Ways to draw same color are 6 (red), 3 (green), and 1 (blue), totaling 10.
    7. (ii) Ways to draw different colors are 36 − 10 = 26.
    8. (ii) Probability is 26/36 which simplifies to 13/18.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    9. (iii) Total outcomes for three coins are 8.
    10. (iii) The condition 'first is heads and exactly two heads' allows only HHT and HTH.
    11. (iii) There are 2 favorable outcomes.
    12. (iii) Probability is 2/8 which simplifies to 1/4.
        why?

        Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    13. (iv) Total four-digit numbers from 1, 2, 3, 4 without repetition are 24.
    14. (iv) For a number to be even, the last digit must be 2 or 4.
    15. (iv) If last digit is 2, there are 3! = 6 numbers; if 4, there are 6 numbers, totaling 12.
    16. (iv) Probability is 12/24 which simplifies to 1/2.
        why?

        Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    17. (v) Total ways to answer 3 questions with 4 options each are 4 × 4 × 4 = 64.
    18. (v) To get exactly 2 correct, choose 2 correct positions in 3 ways and guess the wrong one in 3 ways, giving 9 outcomes.
    19. (v) Probability is 9/64.
        why?

        Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

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

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13. Exercises Q13

    Balls numbered 1 to 4; tree diagrams of the sample space for two draws (i) with the first ball returned, (ii) without returning it; (iii) the sizes of the two sample spaces.

    1. (i) When the first ball is returned, there are 4 choices for the first draw and 4 choices for the second draw.
    2. (i) The total number of outcomes in the sample space is 4 × 4 = 16.
    3. (i) Tree: 4 first branches (1, 2, 3, 4), each splitting into 4 (1, 2, 3, 4): S = {11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44}, where 23 means first 2, then 3.
    4. (ii) When the first ball is not returned, there are 4 choices for the first draw and 3 choices for the second draw.
    5. (ii) The total number of outcomes in the sample space is 4 × 3 = 12.
    6. (ii) Tree: 4 first branches, each splitting into the 3 numbers not yet drawn: S = {12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43}.
    7. (iii) The size of the sample space for drawing with replacement is 16.
    8. (iii) The size of the sample space for drawing without replacement is 12.

    Answer: (i) 16; (ii) 12; (iii) 16 and 12

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14. Exercises Q14

    List the sample space for tossing a coin and drawing one card from six cards numbered 1 to 6.

    1. We toss a coin which can result in Head (H) or Tail (T).
    2. We draw one card from six cards numbered 1 to 6, so the card can be 1, 2, 3, 4, 5, or 6.
    3. The sample space consists of all possible pairs of (Coin outcome, Card number).
    4. The outcomes are: (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6).
    5. Counting these outcomes, we find there are 12 total possibilities.

    Answer: The sample space has 12 outcomes: {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

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15. Exercises Q15

    Three coins are tossed and the number of heads recorded. Which list is the sample space, and why do the others fail: (i) {1, 2, 3}, (ii) {0, 1, 2}, (iii) {0, 1, 2, 3, 4}, (iv) {0, 1, 2, 3}?

    1. When three coins are tossed, the possible outcomes for the number of heads range from 0 to 3.
    2. Option (i) {1, 2, 3} is incorrect because it excludes the case where no heads appear.
    3. Option (ii) {0, 1, 2} is incorrect because it excludes the case where all three coins show heads.
    4. Option (iii) {0, 1, 2, 3, 4} is incorrect because it is impossible to get 4 heads from only 3 coins.
    5. Option (iv) {0, 1, 2, 3} correctly lists all possible counts of heads without any impossible values.

    Answer: (iv) {0, 1, 2, 3}

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16. Exercises Q16

    A drop of dye falls at random on a 3 m × 2 m rectangle containing a circle of diameter 1 m (Fig. 7.8). Probability that it lands inside the circle.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. The circle (diameter 1 m) lies wholly inside the 3 m × 2 m rectangle, whose area is 3 × 2 = 6 m². Every point of the rectangle is equally likely.
    2. The diameter of the circle is 1 m, so its radius is 0.5 m.
    3. The area of the circle is π × (0.5)² = π/4 m².
       why?

       Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

    4. The probability is the ratio of the circle's area to the rectangle's area: (π/4) ÷ 6.
       why?

       Probability of an event = number of outcomes in its favour ÷ total number of equally likely outcomes. It is always from 0 (impossible) to 1 (certain).

    5. Simplifying the fraction gives π/24.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    6. Using π ≈ 3.14, the value is 3.14/24 ≈ 0.13.

    Answer: π/24 ≈ 0.13

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