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A game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at: (i) 8? (ii) An odd number? (iii) A number greater than 2? (iv) A number less than 9? (v) A multiple of 3?

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Sample space has 8 numbers {1, 2, 3, 4, 5, 6, 7, 8}. The probabilities are: (i) P(8) = 1/8, (ii) P(odd) = 4/8 = 1/2, (iii) P(>2) = 6/8 = 3/4, (iv) P(<9) = 8/8 = 1, (v) P(multiple of 3) = 2/8 = 1/4.

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1 Answer

  1. Total possible outcomes = 8, as the arrow can point to {1, 2, 3, 4, 5, 6, 7, 8}.

    (i) Favourable outcome = {8}. Probability = 1/8 = 0.125.

    (ii) Odd numbers = {1, 3, 5, 7}. Probability = 4/8 = 1/2 = 0.5.

    (iii) Numbers greater than 2 = {3, 4, 5, 6, 7, 8}. Probability = 6/8 = 3/4 = 0.75.

    (iv) Numbers less than 9 = {1, 2, 3, 4, 5, 6, 7, 8}. Probability = 8/8 = 1 (certain event).

    (v) Multiples of 3 = {3, 6}. Probability = 2/8 = 1/4 = 0.25.

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/

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