Probability


  1. An urn contains 3 red and 4 green marbles. If three marbles are picked at random, what is the probability that two are green and one is red ?









  1. View Hint View Answer Discuss in Forum

    Number of ways to select 3 marbles out of 7 marbles = n(s) = 7C3 = 35
    Probability that 2 are green and 1 is red = n(E) = 4C2 x 3C1 = 18

    Correct Option: B

    Number of ways to select 3 marbles out of 7 marbles = n(s) = 7C3 = 35
    Probability that 2 are green and 1 is red = n(E) = 4C2 x 3C1 = 18
    ∴ Required probability = 18/35


Direction: Study the given information carefully and answer the question that follow.
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.

  1. If four marbles are picked at random, what is the probability that at least one is blue ?









  1. View Hint View Answer Discuss in Forum

    Probability that none is blue from four marbles
    = 15 - 4C4 / 15C4
    = 11C4 / 15C4

    Correct Option: D

    Probability that none is blue from four marbles
    = 15 - 4C4 / 15C4
    = 11C4 / 15C4 = 22/91

    So, probability that at least one is blue from four marbles = 1 - 22/91 = 69/91



  1. If three marbles are picked at random, what is the probability that two are blue and one is yellow ?









  1. View Hint View Answer Discuss in Forum

    Ways of selection of two blue marbles = 4C2
    Ways of selection of one yellow marble = 3C1
    Ways of selection of three marbles = 15C3
    So, required probability = (4C2 x 3C1) / (15C3)
    = [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]

    Correct Option: C

    Ways of selection of two blue marbles = 4C2
    Ways of selection of one yellow marble = 3C1
    Ways of selection of three marbles = 15C3
    So, required probability = (4C2 x 3C1) / (15C3)
    = [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]
    = [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)]
    = (6 x 3) / (5 x 7 x 13)
    = 18 / 455


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









  1. View Hint View Answer Discuss in Forum

    Total number of marbles = 6 + 4 + 2 + 3 = 15
    Ways of selection of two red marbles = n(E) = 6C2
    Ways of selection of two marbles = n(S) = 15C2

    Correct Option: D

    Total number of marbles = 6 + 4 + 2 + 3 = 15
    Ways of selection of two red marbles = n(E) = 6C2
    Ways of selection of two marbles = n(S) = 15C2
    So, required probability = (6C2) / (15C2)
    = (6 x 5) / (15 x 14) = 1/7



  1. If two marbles are picked at random, what is the probability that either both are green or both are yellow ?









  1. View Hint View Answer Discuss in Forum

    Ways of selection of two green marbles = 2C2 = 1
    Ways of selection of two yellow marbles = 3C2 = 3
    So, probability (both are green) = 1/15C2 ....(I)
    Probability (both are yellow) = 3/15C2 ....(Ii)

    Correct Option: D

    Ways of selection of two green marbles = 2C2 = 1
    Ways of selection of two yellow marbles = 3C2 = 3
    So, probability (both are green) = 1/15C2 ....(I)
    Probability (both are yellow) = 3/15C2 ....(Ii)

    Then, required probability = 1/15C2 + 3/15C2 = 4/ 15C2
    = 4/105