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NCERT Class 9 Mathematics Exercise Set 7.4 Solutions: The Mathematics of Maybe: Introduction to Probability

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise Set 7.4 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.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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  1. Ex 7.4 Q1

    Basket A: 1 apple, 2 oranges; basket B: 1 banana, 1 mango. One fruit is picked from each. (i) Tree diagram of the pairs; (ii) the sample space; (iii) probability of one apple and one banana.

    1. (i) Call the two oranges O₁ and O₂ so that every fruit in basket A is a separate, equally likely choice: Apple, O₁, O₂.
    2. (i) Tree: from the start draw 3 branches (Apple, O₁, O₂); from the end of each draw 2 branches (Banana, Mango). That gives 3 × 2 = 6 paths.
    3. (ii) The 6 outcomes are (Apple, Banana), (Apple, Mango), (O₁, Banana), (O₁, Mango), (O₂, Banana), (O₂, Mango).
    4. (iii) Only one path, (Apple, Banana), has an apple and a banana.
    5. (iii) P(apple and banana) = 1/6.
      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).

    Answer: (i) A tree with 6 paths; (ii) {(Apple, Banana), (Apple, Mango), (O₁, Banana), (O₁, Mango), (O₂, Banana), (O₂, Mango)}; (iii) 1/6

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  2. Ex 7.4 Q2

    A box has 3 red, 4 black and 2 green pens. You pick one and put it back, then a friend does the same. (i) Possible colour outcomes and a tree diagram; (ii) probability that both pens are the same colour.

    1. Total number of pens in the box = 3 + 4 + 2 = 9.
    2. Since the pen is replaced, the total outcomes for the first pick is 9 and for the second pick is also 9.
    3. Total number of possible outcomes = 9 × 9 = 81.
    4. (i) The possible colour outcomes are: (Red, Red), (Red, Black), (Red, Green), (Black, Red), (Black, Black), (Black, Green), (Green, Red), (Green, Black), (Green, Green).
    5. (i) Tree: 3 colour branches (Red, Black, Green) for your pick, each splitting into the same 3 branches for your friend's pick, giving the 9 colour pairs above.
    6. (ii) Number of ways both pens are Red = 3 × 3 = 9.
    7. Number of ways both pens are Black = 4 × 4 = 16.
    8. Number of ways both pens are Green = 2 × 2 = 4.
    9. Total number of favourable outcomes = 9 + 16 + 4 = 29.
    10. Probability that both pens are the same colour = 29/81.
      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) 9 colour outcomes: RR, RB, RG, BR, BB, BG, GR, GB, GG (B = black); (ii) 29/81

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