Quantitative Aptitude: Probability Questions – Set 6

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

  1. Three cards are drawn at random from a well-shuffled deck of cards. What is the probability of drawing a king, a Queen, and a jack?
    A) 16/5525
    B) 64/5225
    C) 16/225
    D) 4/525
    E) None
    View Answer
    Option A
    Solution:

    There are 4 Kings, 4 Queens and 4 Jacks in a pack of cards.
    Probability =(4C1*4C1*4C1)/52C3
    (4*4*4)/(52*51*50/1*2*3)= 16/5525
  2. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?
    A) 1/2
    B) 3/4
    C) 7/4
    D) 3/8
    E) None
    View Answer
    Option B
    Solution:

    Two dice thrown n(s)= 6*6 = 36
    Two no getting product even is = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27
    So Probability = 27/36 = 3/4
  3. A problem is given to three students whose chances of solving it are 1/3, 1/4 and 1/5 respectively. What is the probability that the problem will be solved?
    A) 1/2
    B) 3/4
    C) 3/5
    D) 4/5
    E) None
    View Answer
    Option C
    Solution:

    Given chances of students to solve is 1/3, 1/4, 1/5
    P(the problem will be solved) = 1 – P(none solves the problem)
    P(none solves the problem)=(1-1/3)*(1-1/4)*(1-1/5)
    =2/3*3/4*4/5=24/60
    P(the problem will be solved)=1-24/60
    =(60-24)/60=36/60=3/5
  4. Two Persons attend an interview for two vacancies. The probability of 1st person selection is (1/4) and the probability of 2nd person selection is (1/6). What is the probability that only one of them is selected ?
    A) 1/4
    B) 2/7
    C) 4/5
    D) 1/3
    E) None
    View Answer
    Option D
    Solution:

    Give chance for 1st and 2nd person get selected is 1/4 and 1/6
    1st person not get selected chance is (1-1/4)=3/4
    2nd person not get selected chance is (1-1/6)=5/6
    P(only one get selected)=P(1st selected and 2nd not selected) + P(1st not selected and 2nd selected)
    =(1/4*5/6) + (3/4*1/6) =8/24=1/3
  5. Four different objects 1,2,3 are distributed at random in four places. What is the probability that none of the objects occupies the place corresponding to its number ?
    A) 1/7
    B) 1/2
    C) 2/3
    D) 4/7
    E) None
    View Answer
    Option B
    Solution:

    PLACE: 1 2 3 4
    Ways 2134 , 2143 , 2314 ,2341 , 2413 ,2431
    None of the objects occupies the place of that number = 2143 ,2341,2413
    n(S) = 6, n(E) = 3
    p = 3/6 = 1/2
  6. A bag contain 6 green and 9 yellow flowers. Two successive drawing of four flowers are made such that the flowers are not replaced before the second draw. find the probability that the first draw gives 4 green flowers and second draw gives 4 yellow flowers.
    A) 16/97
    B) 8/453
    C) 3/715
    D) 5/715
    E) None
    View Answer
    Option C
    Solution:

    In 1st draw 4 green flowers selected then P=6C4/15C4
    =15/1365=1/91
    In 2nd draw 4 yellow flowers selected then P=9C4/11C4
    =126/330=21/55
    The required probability=1/91*21/55=3/715
  7. A committee of 4 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls?
    A) 2/3
    B) 1/6
    C) 2/7
    D) 4/7
    E) None
    View Answer
    Option B
    Solution:

    ways 4 girls and 0 boys
    =4C4/9C4
    ways 3 girls and 1 boy
    =4C3*5C1/9C4
    Total probability = 4C4/9C4 + 4C3*5C1/9C4
    =1/126+ 20/126=21/126=1/6
  8. P and Q toss a coin alternately till one of them tosses a tail and wins the game. If P starts the game, find their respective probability of winning.
    A) 1/3, 1/4
    B) 1/4 , 2/3
    C) 2/3, 1/3
    D) 1/3, 1/2
    E) None
    View Answer
    Option C
    Solution:

    P has 1 option to win after Q toses head
    Q has 1 option to win after P toses head
    and then there is the first toss which can grant A a win.
    Hence there are 3 options of winning
    P: 2/3 Q:1/3

Directions (9-10): The odds against a husband who is 45 years old, living till he is 70 are 7:5 and the odds against his wife who is now 36, living till she is 61 are 5:3 . Find the probability that

  1. None of them will be alive 25 years hence.
    A) 35/96
    B) 42/96
    C) 55/96
    D) 35/81
    E) None
    View Answer
    Option A
    Solution:

    Husband: 7:5 [(7/12) against (5/12) infavor)
    Wife : 5:3 [(5/8) against (3/8) infavor)
    (7/12)*(5/8)=35/96
  2. Exactly one of them will be alive 25 years hence :
    A) 29/48
    B) 42/57
    C) 23/57
    D) 23/48
    E) None
    View Answer
    Option D
    Solution:

    Husband: 7:5 [(7/12) against (5/12) infavor)
    Wife : 5:3 [(5/8) against (3/8) infavor)
    (5/12)*(5/8)+(3/8)*(7/12)=23/48

Click here for Probability Topic Questions

 

Related posts

13 Thoughts to “Quantitative Aptitude: Probability Questions – Set 6”

  1. %%%

    mam in question 8th …how Q’s value = 1/3
    game starts with P…if
    P(H) Q(T)
    Q (E) = 1/2 ?????

    1. Premashanthi

      Probability means possibitity na.. there is 3 possibities, in that Q has 1 possibility to win..so it is 1/3

      1. %%%

        isme chances kitne h ye to confirm hi nhi h na……..& as given that game starts with P than there is a probablity also to win in the very 1st chance ….so here
        P (E) = 1/2………….ye bhi to ho skta h na..ki q ka chance hi na aye……..

  2. the walking dead

    7-10:)

  3. Silent Reader

    Plz post latest pattern probablity Ques…..

  4. garima shah

    mam pls explain 8th no…

  5. gomathy priya

    @shubhra mam, here 7:5 means?7 indicates?

    1. 7 indicates in favor of event
      5 – against

      1. gomathy priya

        ty mam:)

  6. kumkum ahuja

    nice quiz mam:)

Leave a Comment