Sunday, May 19, 2024
Banking QuizQuant

Quantitative Aptitude: Probability Set 2

Directions (1-3): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is 2/5. There are 10 blue balls.

  1. What is the probability of choosing one blue ball?
    A) 2/7
    B) 1/4
    C) 1/3
    D) 2/5
    E) 3/7
    View Answer
     Option C
     Solution:
    Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3
  2. What is the total number of balls in the urn?
     A) 45
    B) 34
    C) 40
    D) 30
    E) 42
    View Answer
     Option D
     Solution:
    Probability of choosing one blue ball is 1/3
    And total blue balls are 10. So with 10/30 we get probability as 1/3
    So total balls must be 30
  3. If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 2 or 3?
    A) 3/4
    B) 4/5
    C) 1/4
    D) 1/3
    E) 2/3     
    View Answer
     Option E
     Solution:
    There are 30 balls in the urn.
    Multiples of 2 up to 30 = 30/2 = 15
    Multiples of 3 up to 30 = 30/3 = 10 (take only whole number before the decimal part)
    Multiples of 6 (2×3) up to 30 = 30/6 = 5
    So total such numbers are = 15 + 10 – 5 = 20
    So required probability = 20/30 = 2/3     
  4. There are 2 brothers A and B. Probability that A will pass in exam is 3/5 and that B will pass in exam is 5/8. What will be the probability that only one will pass in the exam?
    A) 12/43
    B) 19/40
    C) 14/33
    D) 21/40
    E) 9/20
    View Answer
     Option B
     Solution:
    Only one will pass means the other will fail
    Probability that A will pass in exam is 3/5. So Probability that A will fail in exam is 1 – 3/5 = 2/5
    Probability that B will pass in exam is 5/8. So Probability that B will fail in exam is 1 – 5/8 = 3/8
    So required probability = P(A will pass)*P(B will fail) + P(B will pass)*P(A will fail)
    So probability that only one will pass in the exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40   
  5. If three dices are thrown simultaneously, what is the probability of having a same number on all dices?
    A) 1/36
    B) 5/36
    C) 23/216
    D) 1/108
    E) 17/216
    View Answer
     Option A
     Solution:
    Total events will be 6*6*6 = 216
    Favorable events for having same number is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4}, {5,5,5}, {6,6,6} – so 6 events
    Probability of same number on all dices is 6/216 = 1/36    
  6. There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?
    A) 52/125
    B) 53/150
    C) 17/50
    D) 37/150
    E) 32/75     
    View Answer
     Option E
     Solution:
    Multiples of 3 up to 150 = 150/3 = 50
    Multiples of 7 up to 150 = 150/7 = 21 (take only whole number before the decimal part)
    Multiples of 21 (3×7) up to 150 = 150/21 = 7  
    So total such numbers are = 50 + 21 – 7 = 64
    So required probability = 64/150 = 32/75     
  7. There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a ticket which has a prime number on it?
    A) 3/55
    B) 5/58
    C) 8/21
    D) 16/55
    E) 4/13
    View Answer
     Option D
     Solution:
    Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 41, 47, 53.
    So probability = 16/55
  8. A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue?
    A) 61/110
    B) 59/108
    C) 45/134
    D) 53/108
    E) 57/110
    View Answer
     Option D
     Solution:
    Case 1: Ball from first bag is white, from another is blue
    So probability = 4/9 * 7/12 = 28/108
    Case 1: Ball from first bag is blue, from another is white
    So probability = 5/9 * 5/12 = 25/108
    Add the cases
    So required probability = 28/108 + 25/108 = 53/108
  9. The odds against an event are 2 : 3 and the odds in favor of another independent event are 3 : 4. Find the probability that at least one of the two events will occur.
    A) 11/35
    B) 27/35
    C) 13/35
    D) 22/35
    E) 18/35
    View Answer
     Option B
     Solution:
    Let 2 events A and B
    Odds against A are 2 : 3
    So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5
    Odds in favor of B are 3 : 4
    So probability of occurrence of B = 3/(3+4) = 3/7. And non-occurrence of B = 4/7
    Probability that at least one occurs 
    Case 1: A occurs and B does not occur
    So probability = 3/5 * 4/7 = 12/35
    Case 2: B occurs and A does not occur
    So probability = 3/7 * 2/5 = 6/35
    Case 3: Both A and B occur
    So probability = 3/5 * 3/7 = 9/35
    So probability that at least 1 will occur = 12/35 + 6/35 + 9/35 = 27/35    
  10. The odds against an event are 1 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.
    A) 17/28
    B) 5/14
    C) 11/25
    D) 9/14
    E) 19/28
    View Answer
     Option A
     Solution:
    Let 2 events A and B
    Odds against A are 1 : 3
    So probability of occurrence of A = 3/(1+3) = 3/4. And non-occurrence of A = 1/4
    Odds in favor of B are 2 : 5
    So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7
    Case 1: A occurs and B does not occur
    So probability = 3/4 * 5/7 = 15/28
    Case 2: B occurs and A does not occur
    So probability = 2/7 * 1/4 = 2/28
    So probability that one will occur = 15/28 + 2/28 = 17/28

 

 

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