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Probability Test - 6

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Probability Test - 6
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  • Question 1
    1 / -0
    In a collection of 6 English books and 4 Reasoning books, the probability that 3 particular English books will be lying together is
    Solution
    Number of ways of 3 particular English books will be lying together = 8! x 3!

    Total number of ways = (6 + 4)! = 10!

    Required probability =
  • Question 2
    1 / -0
    You are given a box with 25 cards in it. 10 of these cards have letter "I" printed on them and the other 15 have the letter "T" printed on them. If you pick up 3 cards one by one at random and keep them in the same order, the probability of making the word I.I.T. is
    Solution
    Required Probability,
    P('I' first Draw)+P('I' first Draw) + P('T' first Draw)
    = [10C1]/[25C1] + [9C1]/[24C1] + [15C1]/[23C1]
    =[10/25] × [9/24] × [15/23]
    = 1350/13800
    = 45/460
    = 9/92
  • Question 3
    1 / -0
    N students have to stand in a row. If all possible permutations are equally likely, the probability of two particular students standing side by side is
    Solution
    n (S) = number of ways of arranging N students in a row = NPN = N!
    Consider two particular students as one.
    Then (N – 1) students can be arranged in (N – 1)! ways and these two students can be arranged amongst
    themselves in 2! ways.
    n (E) = (N – 1)! ´ 2!
    Hence, required probability =
  • Question 4
    1 / -0
    The odds against an event A are 5 : 3 and odds in favour of another independent event B are 7 : 5. The chances that neither A nor B occurs is
    Solution
    The odds against an event A are 5 : 3.
    So, Probability of occurence if event A =

    And odds in favour of another independent event B are 7 : 5.
    So, Probability of occurence if event B =

    The chances that neither A nor B occurs is =
  • Question 5
    1 / -0
    A bag contains 6 white and 4 black balls. A man pulls out 2 balls at random. The probability that they are of the different colour is
    Solution
    n (S) = number of ways of pulling 2 balls out of 10
    = 10C2 = = 45.
    n (E) = (number of ways of pulling 2 balls out of 6 or 2 balls out of 4)
    = (6C1 x 4C1) = (6 x 4) = 24.
    Required probability = =
  • Question 6
    1 / -0
    A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade. The chance that he will fail in the first two draws is
    Solution
    Required probability =
  • Question 7
    1 / -0
    The probability of getting a sum of more than 7 when a pair of dice is thrown, is
    Solution
    Favourable Cases = (2, 6) (3, 5) (3, 6) (4, 4) (4, 5) (4, 6) (5, 3) (5, 4) (5, 5) (5, 6)
    (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) = 15 in total
    Required probability = = .
  • Question 8
    1 / -0
    From each of two married couples, one of the partners is selected randomly. The probability that those selected are of the opposite sex is
    Solution
    number of favourable outcomes/ total number of outcomes
    = ½
  • Question 9
    1 / -0
    A bag contains 15 white and 4 green balls. One ball is drawn at random. The probability that it is green is
    Solution
    Total number of balls = 15 + 4 = 19
    Number of green balls = 4
    Required probability =
  • Question 10
    1 / -0
    Two fair dice are thrown. Let X be the event that the first die shows an even number and Y be the event that the second die shows an odd number. The two events X and Y are
    Solution
    Events are independent but mutually exclusive as the outcomes of both events are different.
  • Question 11
    1 / -0
    The probability of having 53 Mondays in a year is
    Solution
    Normal year has 365 days.
    365/7 = 52, remainder = 1
    52 weeks are there 52 Sundays, 1 day extra;
    For 53 Sundays this day should be a Sunday. So, P (53 Sundays) = 1/7.
  • Question 12
    1 / -0
    The probability that A can solve a problem is 1/3, and the probability that B can solve the problem is 1/4. The probability that at least one of A and B will be able to solve the problem is
    Solution
  • Question 13
    1 / -0
    Two cards are drawn from a pack. The probability of getting a king and a queen is
    Solution
    Probability of getting a king and a queen = (K Q) (Q K) = P(KQ) + P(QK)
    = =
  • Question 14
    1 / -0
    A husband and wife appear for an interview for the same post. The probability of husband's selection is and that of wife's is . What is the probability that none of them will be selected?
    Solution
    The probability of husband's selection is .

    The probability of wife selection is .

    The required probability =

    = =
  • Question 15
    1 / -0
    There are 40 cards with numbers 1 to 40 printed on them. What is the probability of finding a card with a multiple of 4 in a draw?
    Solution
    Total number of possible outcomes = 40

    Favourable outcomes = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 = 10

    So, the required probability = .


  • Question 16
    1 / -0
    The chances of a win of two race horses are and respectively. What is the probability that at least one will win when the horses are running in different races?
    Solution
  • Question 17
    1 / -0
    Four letters are written to different persons and addresses on the envelopes are also written. Without looking at the addresses, the probability that letters go into right envelopes is
    Solution
    There are four letters and four directed envelopes.
    Therefore, they can be put into the envelopes in 4C4 = 4 ! = 624 ways, out of which only one is correct.
    Thus, the required probability = 1/24
  • Question 18
    1 / -0
    A work is given to 4 men and their probabilities for doing the work are , , and respectively. The probability that the work done is
  • Question 19
    1 / -0
    Ram and Ravana throw a dice. The chance that both Ram and Ravana throw the same number is
    Solution
    Required probability = 6/36 = 1/6.
  • Question 20
    1 / -0
    If A and B are two events such that
    (i) P(A B) = , P(B A) = .
    (ii) P(A) = , then the events A and B are
    Solution
    As, Occurance of A does not affect the probability of B occurring. So, two events, A and B, are independent.
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