Quantitative Aptitude: Probability Questions – Set 4

Probability Questions for Bank PO Exams – SBI PO, NIACL, NICL, BoB PO, IBPS PO, and other exams

  1. A bag contains 6 red, 2 blue and 4 green balls. 3 balls are chosen at random. What is the probability that at least 2 balls chosen will be red?
    A) 2/7
    B) 1/2
    C) 1/3
    D) 2/5
    E) 3/7
    View Answer
    Option B
    Solution:
    There will be 2 cases
    Case 1: 2 red, 1 blue orgreen
    Prob. = 6C2 × 6C1 / 12C3 = 9/22
    Case 2: all 3 red
    Prob. = 6C3 / 12C3 = 2/22
    Add the cases, required prob. = 9/22 + 2/22 = 11/22 = 1/2
  2. Tickets numbered 1 to 250 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 4 or 7?
    A) 83/250
    B) 89/250
    C) 77/250
    D) 93/250
    E) 103/250
    View Answer
    Option B
    Solution:
    Multiples of 4 up to 120 = 250/4 = 62
    Multiples of 7 up to 120 = 250/7 = 35 (take only whole number before the decimal part)
    Multiple of 28 (4×7) up to 250 = 250/28 = 8
    So total such numbers are = 62 + 35 – 8 = 89
    So required probability = 89/250  
  3. From a deck of 52 cards, 3 cards are chosen at random. What is the probability that all are face cards?
    A) 14/1105
    B) 19/1105
    C) 23/1105
    D) 11/1105
    E) 26/1105
    View Answer
    Option D
    Solution:
    There are 3*4 = 12 face cards in 52 cards
    So required probability = 12C3 / 52C3 = 11/1105
  4. One 5 letter word is to be formed taking all letters – S, A, P, T and E. What is the probability that this the word formed will contain all vowels together?
    A) 2/5
    B) 3/10
    C) 7/12
    D) 3/5
    E) 5/12
    View Answer
    Option A
    Solution:
    Total words that can be formed is 5! = 120
    Now vowels together:
    Take: S, P, T and AE
    So their arrangement is 4! * 2! = 48
    So required probability = 48/120 = 2/5
  5. One 5-digit number is to be formed from numbers – 0, 1, 3, 5, and 6 (repetition not allowed). What is the probability that number formed will be even?
    A) 8/15
    B) 7/16
    C) 7/15
    D) 3/10
    E) 13/21
    View Answer
    Option B
    Solution:
    Two cases:
    Case 1: 0 at last place
    So 4 choices for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 4*3*2*1 = 24
    Case 2: 6 at last place
    For 5-digit number 0 cannot be placed at 1st place or cannot be 1st digit
    So 3 choices (1, 3, 5) for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 3*3*2*1 = 18
    So total choices = 24+18 = 42
    Number total 5-digit numbers that can be formed from 0, 1, 3, 5, and 6
    0 not allowed at 1st place, so 4 choices for 1st place, 4 for 2nd, 3 for 3rd, 2 for 4th and 1 for 5th. Sp total = 4*4*3*2*1 = 96
    So required probability = 42/96 = 7/16

Directions (6-8): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains:
24 green balls. Red balls are 4 more than blue balls. Probability of selecting 1 red ball is 4/13

Bag 2 contains:
Total balls are 8 more than 7/13 of balls in bag 1. Probability of selecting 1 red ball is 1/3. The ratio of green balls to blue balls is 1 : 2

Bag 3 contains:
Red balls equal total number of green and blue balls in bag 2. Green balls equal total number of green and red balls in bag 2. Probability of selecting 1 blue ball is 3/14.

  1. 1 ball each is chosen from bag 1 and bag 2, What is the probability that 1 is red and other blue?
    A) 15/128
    B) 21/115
    C) 17/135
    D) 25/117
    E) 16/109
    View Answer
    Option D
    Solution:
    Let red = x, so blue = x-4
    So
    x/(24+x+(x-4)) = 4/13
    Solve, x = 16
    So bag 1: red = 16, green  = 24, blue = 12
    NEXT:
    bag 2: total = 8 + 7/13 * 52 = 36
    green and blue = y and 2y. Let red balls = z
    So z + y + 2y = 36…………………(1)
    Now Prob. of red = 1/3
    So z/36 = 1/3
    Solve, z = 12
    From (1), y = 8
    So bag 2: red = 12, green  = 8, blue = 16
    NEXT:
    bag 3: red = 8+16 = 24, green = 12+8 = 20
    Blue prob. = 3/14
    So a/(24+20+a) = 3/14
    Solve, a = 12
    So bag 3: red = 24, green  = 20, blue = 12
    Now probability that 1 is red and other blue::
    16/52 * 16/36 + 12/52 * 12/36 = 25/117
  2. Some green balls are transferred from bag 1 to bag 3. Now probability of choosing a blue ball from bag 3 becomes 3/16. Find the number of remaining balls in bag 1.
    A) 60
    B) 58
    C) 52
    D) 48
    E) 44
    View Answer
    Option E
    Solution:
    blue balls in bag 3 are 12
    Let x green balls are transferred. So
    12/(56+x) = 3/16   [56 was number of bags in bag 3 before transfer] Solve, x = 8
    So remaining number of balls in bag 1 = 52-8 = 44
  3. Green balls in ratio 4 : 1 from bags 1 and 3 respectively are transferred to bag 4. Also 4 and 8 red balls from bags 1 and 3 respectively . Now probability of choosing green ball from bag 4 is 5/11. Find the number of green balls in bag 4?
    A) 12
    B) 15
    C) 10
    D) 9
    E) 11
    View Answer
    Option C
    Solution:
    4x and x = 5x green balls
    4+8 = 12 red balls
    So 5x/(5x+12) = 5/11
    Solve, x = 2
    5*2 = 10 green balls

Directions (9-10): There are 3 people – A, B and C. Probability that A speaks truth is 3/10, probability that B speaks truth is 3/7 and probability that C speaks truth is 5/6. For a particular question asked, at most 2 people speak truth. All people answer to a particular question asked.

  1. What is the probability that B will speak truth for a particular question asked?
    A) 7/18
    B) 14/33
    C) 4/15
    D) 9/28
    E) 10/33
    View Answer
    Option D
    Solution:
    In any case B speaks truth. Now at most 2 people speak truth for 1 question
    So case 1: B and A speaks truth
    Probability = 3/7 * 3/10 * (1-5/6) = 3/140
    Case 2: B and C speaks truth
    Probability = 3/7 * ( 1- 3/10) * 5/6 = 5/20
    Case 3: Only B speaks truth
    Probability = 3/7 * ( 1- 3/10) * (1-5/6) = 1/20
    Add the three cases = 6/20 + 3/140 = 45/140 = 9/28
  2. A speaks truth only when B does not speak truth, then what is the probability that C does not speak truth on a question?
    A) 11/140
    B) 21/180
    C) 22/170
    D) 13/140
    E) None of these
    View Answer
    Option A
    Solution:
    Case 1: B does not speak truth, A speaks truth
    So A speaks truth here
    Probability that C does not speak truth = 3/10 * (1 – 3/7) * ( 1- 5/6) = 1/35
    Case 2: B speaks truth
    So A does not speak truth here
    Probability that C does not speak truth = ( 1- 3/10) * 3/7 * ( 1- 5/6) = 1/20
    So total = 1/35 + 1/20 = 11/140

 

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