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

9. 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
   
10. 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