Probability

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  • Q1

    The probability that in a group of n persons all will have their birthdays on different days of the year is:

    Marks:1
    Answer:


    Explanation:

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  • Q2

    Cards are drawn one by one without replacement from a pack of 52 cards. The probability that 10 cards will precede the first ace is

    Marks:1
    Answer:

    492/12495

    Explanation:

    Total number of cards = 52
    Number of aces = 4
    Probability of one ace = 4/52
    Probability of othere ten cards taken
    one by one without replacement = (48/51)(47/50)(46/49)...(39/42)
    Therefore,
    Required Probability =
    (48/51)(47/50)(46/49)...(39/42).(4/52)
                                      = 492/12495

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  • Q3

    In 50 tickets of a lottery there are three prizes first, second and third. The probability that a person having 5 tickets will win a price is

    Marks:1
    Answer:

    Explanation:

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  • Q4

    The probability of getting 10 in a single throw of three fair dice is

    Marks:1
    Answer:

    1/8.

    Explanation:

    Exhaustive no. of cases = 63

    10 can appear on three dice either as distinct number as following:

    (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and  each can occur in 3! ways.

    Or 10 can appear on three dice as repeated digits as following :

    (2, 2, 6), (2, 4, 4), (3, 3, 4) and each can occur in  3!/ 2! ways.

      No. of favourable cases  = 33!+3 (3!/ 2!) =27

      Hence, the required probability= 27/ (63) = 1/8.

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  • Q5

    Two dice are thrown simultaneously. The probability of obtaining a total score of seven is

    Marks:1
    Answer:

    1/6.

    Explanation:

    When two dice are thrown then there are 6 × 6 exhaustive cases  n = 36.

    Let A denote the event “total score of 7” when 2 dice are thrown then

          A = [(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1)].

    Thus there are 6 favourable cases.

     m = 6

    By definition P(A) = m/n

          P(A) = 6/36 = 1/6

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