In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up…

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In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

  1. A.

    1/3

  2. B.

    3/4

  3. C.

    7/19

  4. D.

    2/3

Attempted by 9 students.

Show answer & explanation

Correct answer: A

Concept: For equally likely outcomes, the probability of an event E is P(E) = n(E) / n(S), where n(S) is the total number of possible outcomes (the sample space) and n(E) is the number of outcomes favourable to E.

  1. Total balls in the box (the sample space) = 8 red + 7 blue + 6 green = 21.

  2. The event 'neither red nor green' is satisfied only by the blue balls, so the number of favourable outcomes = 7.

  3. Applying the formula: P(neither red nor green) = 7 / 21 = 1/3.

Cross-check: The complementary event 'red or green' has (8 + 6) = 14 favourable outcomes, giving P(red or green) = 14/21 = 2/3. Since the two events partition the sample space, their probabilities must add to 1, and indeed 1/3 + 2/3 = 1, confirming the result.

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