Dear Aspirants, Probability questions are a part of most of the bank and other competitive exams. Here we have provided you with the best practice questions for probability in quiz as well as PDF format. Probability Questions For Bank Clerk Prelims Exam will be asked as the application sums. Solving the prelims level probability questions requires good understanding of the basics and easy mind math tricks. Probability questions for bank exams pdf covers all the types of questions asked in the bank prelims examination. Probability is nothing but the possibility of an event occurring. The likelihood of an event ranges from 0 to 1, where 0 means the event to be an impossible one, and one represents the event to be certain. Understand the basic concepts and apply the necessary formulas by practicing the Probability Questions For Bank Clerk Prelims Exam pdf. The common type of probability questions asked in the examination is tossing a coin, choosing a thing and rolling a dice. Based on sample spaces, candidates should solve a question.
The detailed explanation of the solutions are provided in the Probability Questions For Bank Clerk Prelims Exam pdf provided here. In this section of quantitative aptitude, we will learn about the concept of probability keeping in mind the type and level of questions that are asked in all the crucial competitive exams. Solving Probability Questions For Bank Clerk Prelims Exam pdf requires good understanding of the basic concepts. Candidates should analyse the solutions given and understand the tips and shortcuts provided here to solve the Probability Questions For Bank Clerk Prelims Exam pdf. Enhance your preparation in the quantitative aptitude section by utilizing this Probability Questions For Bank Clerk Prelims Exam pdf provided here.
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1) A bag contains 5 yellow, 3 blue and 4 white shirts. 2 shirts are taken out randomly. What is the probability that one shirt is yellow and other shirt is blue?
A.6/23
B.3/22
C.5/22
D.2/11
E.4/33
2) A bag contains 8 balls out which x red balls and remaining are yellow balls. If two balls are drawn at random and the probability that both being red balls is 5/14, then find the number of yellow balls in the bag?
A.1
B.3
C.2
D.4
E.6
3) A box contains red and green balls and the total number of balls in the box is 15. If the probability of picking a green ball is 0.4, find the number of red balls from the box?
A.8
B.7
C.9
D.6
E.12
4) A bag contains (x – 1) red balls, 3 blue balls and 4 green balls. If one ball is taken out randomly and the probability of getting a green ball is 2/5, find the value of x?
A.4
B.6
C.2
D.3
E.1
5) A box contains 6 Pink balls, 3 black balls, 4 orange balls and 2 white balls. If four balls are drawn out randomly, then the probability of getting balls of different colour?
A.11/135
B.48/455
C.23/479
D.34/477
E.None of these
6) A box contains 5 white balls, 6 violet balls and 3 yellow balls. If 2 balls are drawn randomly, then the probability of getting both the balls are either white or violet balls?
A.11/56
B.23/72
C.35/83
D.25/91
E.None of these
7) A bag contains 2 red balls, 4 green balls and x white balls. Two balls are drawn at random and the probability of that balls being green is 2/7, then find the value of x?
A.1
B.2
C.3
D.4
E.None of these
8) A bag contains 5 red, 6 blue and some yellow balls. Probability of drawing one red ball from the bag is 1/3. Three balls are drawn at random. Find the probability that all the balls are of same color?
A.29/413
B.34/455
C.31/451
D.37/453
E.None of these
9) A bag contains 60 balls marked from 1 to 60. If one ball is drawn at random, then what is the probability that is marked with a number divisible by 7?
A.1/15
B.2/15
C.1/5
D.4/15
E.None of these
10) A box contains 3 apples and 2 Oranges. If two fruits are picked at random, then what is the probability that is at least one apple in the drawn fruits?
A.9/10
B.4/5
C.7/10
D.3/5
E.None of these
Answers :
1) Answer: C
Total number of shirts = 5 + 3 + 4 = 12
No. of ways of drawing 2 shirts = 12C2Â = (12*11)/2 = 66
No. of ways of drawing 1 yellow and 1 blue shirt = (5C1*3C1) = 5*3 = 15
Probability = 15/66 = 5/22
2) Answer: B
Number of red balls=x
Number of yellow balls=(8 – x)
5/14=xC2/8C2
5/14=x * (x – 1)/56
20=x2 – x
x2 – x – 20=0
x2 – 5x + 4x – 20=0
x(x – 5) + 4(x – 5)=0
x=-4, 5(negative value neglected)
x=5
Number of yellow balls=8 – 5=3
3) Answer: C
Total number of balls = 15
Number of green balls =x
Number of red balls = 15 – x
xC1Â / 15C1Â = 0.4
x = 6
Number of red balls = 15 – 6 = 9
4) Answer: A
Total number of balls = x – 1 + 4 + 3 = x + 6
2/5 =Â 4C1/(x + 6)C1
2/5 = 4/(x + 6)
20 = 2x + 12
2x = 8
x = 4
5) Answer: B
Total probability n(S) = 15C4Â = (15*14*13*12) / (1*2*3*4)
Required probability n(E) = 6C1Â and 3C1Â and 4C1Â and 2C1
P(E) = n(E)/n(S)
P(E) = (6C1Â and 3C1Â and 4C1Â and 2C1) / 15C4
P(E) = 48/455
6) Answer: D
Total probability n(S) = 14C2
Required probability n(E) = 5C2Â or 6C2
P(E) = n(E)/n(S) = 5C2Â or 6C2Â / 14C2
= > 25/91
7) Answer: A
2/7 = 4C2Â / (x + 6)C2
2/7 = 12/ (x + 6) * (x + 5)
42 = (x + 6) * (x + 5)
42 = x2Â + 5x + 6x + 30
x2 + 11x – 12 = 0
x2 + 12x – x – 12 = 0
x(x + 12) – 1(x + 12) = 0
(x – 1) (x + 12) = 0
x = 1, -12 (negative value is neglect)
x = 1
8) Answer: B
Red = 5
Blue = 6
Let, yellow = n
Total number of balls = 5 + 6 + n = (11 + n)
According to the question
5/(11 + n) = 1/3
= > 15 = 11 + n
= > n = 15 – 11
= > n = 4
Required probability = (5c3+Â 6c3Â +Â 4c3)/15c3
= > (10 + 20 + 4)/455
= > 34/455
9) Answer: B
Number of marked balls divisible by 7 = 7, 14, 21, 28, 35, 42, 49, 56
Required probability = 8/60 = 2/15
10) Answer: A
Required probability = (3C1Â * 2C1Â + 3C2)/5C2
= [(3 * 2) + 3]/(5 * 2)
= 9/10
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