Quantitative Aptitude: Probability – Set 2 (for SBI PO)

In SBI PO Data Interpretation and Analysis section, some other topics are also asked. One of them is Probability which remain easy and will help gain your marks. Probability SBI PO.

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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15 Thoughts to “Quantitative Aptitude: Probability – Set 2 (for SBI PO)”

  1. Chetan

    InQ3)…probality of 2 and 3 ya 2 or 3…pls check..

      1. Blak

        I was wrong, sorry

  2. Blak

    Q7. there are 16 prime numbers, 41 is missing

      1. garima shah

        hello mam ye odds against event nd in favour events ka kya mean h….these r new terms for me..pls bta dijiye………

        1. Odds means chances
          So chances of occurring of event

          Odds in favor = no. of favorable choices : no. of unfavorable choices
          Odds against = no. of unfavorable choices : no. of favorable choices

  3. Dream BIG :))Pappa's Dream❤

    @s@Shubhra_AspirantsZone:disqus di..
    Cant we do questn 8 like dis?
    4C1 x7C1 + 5C1 × 5C1/21C2??

    1. Dream BIG :))Pappa's Dream❤

      It would be like dis..
      Aa gya?
      4c1x7c1/11c2 + 5c1x5c1/10c2

  4. the walking dead

    🙂

  5. kumkum ahuja

    ty 🙂

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