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Direction: Study the following information carefully to answer the question the question that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.

  1. If two caps are picked at random, what is the probability that both are blue?








  1. View Hint View Answer Discuss in Forum

    Total number of caps = 2 + 4 + 5 + 1 = 12
    Total number of outcomes = n(S) = 12 C2 = 66
    Favorable number of outcomes = n(E) = 2C2 = 1

    Correct Option: D

    Total number of caps = 2 + 4 + 5 + 1 = 12
    Total number of outcomes = n(S) = 12 C2 = 66
    Favorable number of outcomes = n(E) = 2C2 = 1
    ∴ Required probability = 1/66

  1. If four caps are picked at random, what is the probability that none is green ?








  1. View Hint View Answer Discuss in Forum

    Total number of caps = 12
    Total number of result = n(S) = 12C4 = 495
    Out of 5 caps, number of ways to not pick a green cap = n(E1) = 5C0 = 1
    and out of 7 caps, number of ways to pic 4 caps = n(E) = 7C4 = 35

    Correct Option: A

    Total number of caps = 12
    Total number of result = n(S) = 12C4 = 495
    Out of 5 caps, number of ways to not pick a green cap = n(E1) = 5C0 = 1
    and out of 7 caps, number of ways to pic 4 caps = n(E) = 7C4 = 35
    ∴ Required probability = (1 x 35)/495 = 7/99

  1. If three caps are picked at random, what is the probability that two are red and one is green ?








  1. View Hint View Answer Discuss in Forum

    Total number of caps = 12
    ∴ n(S) = 12C3 = 220
    n(E1) = Out of 4 red caps, number of ways to pick 2 caps = 4C2 = 6
    n(E2) = Out of 5 green caps
    Number of ways to pick one cap = 5C1 = 5
    P(E) = n(E1) x n(E2)/n(S)

    Correct Option: D

    Total number of caps = 12
    ∴ n(S) = 12C3 = 220
    n(E1) = Out of 4 red caps, number of ways to pick 2 caps = 4C2 = 6
    n(E2) = Out of 5 green caps
    Number of ways to pick one cap = 5C1 = 5
    P(E) = n(E1) x n(E2)/n(S) = (6 x 5)/220 = 3/22

  1. If two caps are picked at random, what is the probability that atleast one is red ?








  1. View Hint View Answer Discuss in Forum

    Required probability = 1 - 8C2/ 12C2

    Correct Option: C

    Required probability = 1 - 8C2/ 12C2
    = 1 - 28/66 = 38/66 = 19/33

Direction: Study the given information carefully and answer the questions that follow.
A basket contains 4 red, 5 blue and 3 green marbles.

  1. If two marbles are drawn at random, what is the probability that both are red ?








  1. View Hint View Answer Discuss in Forum

    Total number of possible outcomes
    = n(S) = 12C2 = (12 x 11)/2 = 66

    Total number of favorable events
    = n(E) = 4C2
    = (4 x 3) / (2 x 1) = 6

    Correct Option: D

    Total number of possible outcomes
    = n(S) = 12C2 = (12 x 11)/2 = 66

    Total number of favorable events
    = n(E) = 4C2
    = (4 x 3) / (2 x 1) = 6
    ∴ Required probability = 6/66 = 1/11