Probability


  1. Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 or 7 ?









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    Clearly, n(S) = 20 and E = { 3, 6, 9, 12, 15, 18, 7, 14 } i.e., n(E) = 8

    ∴ P(E) = n(E)/ n(S) = 8/20 = 2/5

    Correct Option: C

    Clearly, n(S) = 20 and E = { 3, 6, 9, 12, 15, 18, 7, 14 } i.e., n(E) = 8

    ∴ P(E) = n(E)/ n(S) = 8/20 = 2/5


  1. The odds against the occurrence of an event are 5 : 4. The probability of its occurrence is ?









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    Number of cases favourable of E = 4
    Total Number of cases = (5 + 4 ) = 9
    ∴ P(E) = 4/9

    Correct Option: B

    Number of cases favourable of E = 4
    Total Number of cases = (5 + 4 ) = 9
    ∴ P(E) = 4/9



  1. A bag contains 8 red and 5 white balls. 2 balls are drawn at random. What is the probability that both are white ?









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    n(S) = Number of ways of drawing 2 balls out of 13 = 13C2 = 13 x 12 / 2= 78
    n(E) = No. of ways of drawing 2 balls out of 5 = 5C2 = 5 x 4 / 2 = 10

    ∴ P(E) = n(E)/n(S) = 10/78 = 5/39

    Correct Option: D

    n(S) = Number of ways of drawing 2 balls out of 13 = 13C2 = 13 x 12 / 2= 78
    n(E) = No. of ways of drawing 2 balls out of 5 = 5C2 = 5 x 4 / 2 = 10

    ∴ P(E) = n(E)/n(S) = 10/78 = 5/39


  1. A bag contains 6 black balls and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white ?









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    Total no. of balls = (6 + 8) = 14
    No. of white balls = 8
    ∴ P(drawing a white ball) = 8/14 = 4/7

    Correct Option: A

    Total no. of balls = (6 + 8) = 14
    No. of white balls = 8
    ∴ P(drawing a white ball) = 8/14 = 4/7



  1. Three unbiased coins are tossed. What is the probability of getting at most 2 heads ?









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    ∵ n(S) = (2)3 = 8
    E = Event of getting 0, or 1 or 2 heads
    = {TTT, TTH, THT, HTT, HHT, HTH, THH}
    ⇒ n(E) = 7

    Correct Option: C

    ∵ n(S) = (2)3 = 8
    E = Event of getting 0, or 1 or 2 heads
    = {TTT, TTH, THT, HTT, HHT, HTH, THH}
    ⇒ n(E) = 7
    ∴ P(E) = n(E)/n(S) = 7/8