Probability Questions For IBPS Clerk

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probability questions for ibps clerk
probability questions for ibps clerk

Probability Questions For IBPS Clerk

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Instructions

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.

Question 1: If four marbles are picked at random, what is the probability that at least one is blue ?

a) 4/11

b) 69/91

c) 11/15

d) 22/91

e) None of these

Question 2: If two marbles are picked at random, what is the probability that both are red ?

a) 1/6

b) 1/3

c) 2/15

d) 2/5

e) None of these

Question 3: If three marbles are picked at random, what is the probability that two are blue and one is yellow?

a) 3/91

b) 1/5

c) 18/455

d) 7/15

e) None of these

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Question 4: If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?

a) 24/455

b) 13/455

c) 11/15

d) 7/91

e) None of these

Question 5: If two marbles are picked at random, what is the probability that either both are green or both are yellow ?

a) 5/91

b) 1/35

c) 1/3

d) 4/105

e) None of these

Question 6: In how many different ways can the letters of the word ‘STRESS’ be arranged

a) 360

b) 240

c) 720

d) 120

e) None of these

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Question 7: A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow?

a) 5/33

b) 5/22

c) 7/22

d) 7/33

e) 7/66

Question 8: A die is thrown twice. What is the probability of getting a sum 7 from the two throws?

a) 5/18

b) 1/18

c) 1/9

d) 1/6

e) 5/36

InstructionsInstructions: Study the given information carefully to answer the questions that follows.

An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.

Question 9: If two marbles are drawn at random, what is the probability that both are red or at least one is red?

a) $\frac{26}{91}$

b) $\frac{1}{7}$

c) $\frac{199}{364}$

d) $\frac{133}{191}$

e) None of these

Question 10: If three marbles are drawn at random, what is the probability that at least one is yellow?

a) $\frac{1}{3}$

b) $\frac{199}{364}$

c) $\frac{165}{364}$

d) $\frac{3}{11}$

e) None of these

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Question 11: If eight marbles are drawn at random, what is the probability that there are equal numbers of marbles of each colour?

a) $\frac{4}{7}$

b) $\frac{361}{728}$

c) $\frac{60}{1001}$

d) $\frac{1}{1}$

e) None of these

Question 12: If three marbles are drawn at random, what is the probability that none is green?

a) $\frac{2}{7}$

b) $\frac{253}{728}$

c) $\frac{10}{21}$

d) $\frac{14}{91}$

e) $\frac{30}{91}$

Question 13: If four marbles are drawn at random, what is the probability that two are blue and two are red?

a) $\frac{10}{1001}$

b) $\frac{9}{14}$

c) $\frac{17}{364}$

d) $\frac{2}{7}$

e) None of these

Question 14: In how many different ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together?

a) 720

b) 1440

c) 5040

d) 3600

e) 4800

Question 15: A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?

a) $\frac{41}{190}$

b) $\frac{21}{190}$

c) $\frac{59}{290}$

d) $\frac{99}{190}$

e) $\frac{77}{190}$

Question 16: If 6 boys and 6 girls have to sit in a round circular music chair. So, that there is a girl between every 2 boys. Find the number of ways they can sit?

a) 6! × 5!

b) 6! × 4!

c) 6! × 3!

d) 6! × 2!

e) None of these

Question 17: What is the number of words formed from the letters of the word ‘JOKE’ So that the vowels and consonants alternate?

a) 4

b) 8

c) 12

d) 18

e) None of these

Question 18: In how many different ways can 5 men and 3 women be seated in a row such that no two women are next to each other?

a) 12200

b) 14400

c) 15600

d) 16400

e) None of the above

Question 19: In how many different ways can the letter of the word ‘SIMPLE’ be arranged ?

a) 520

b) 120

c) 5040

d) 270

e) None of these

Question 20: In how many different ways can the letters of the word ‘SECOND’ be arranged ?

a) 720

b) 120

c) 5040

d) 270

e) None of these

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Answers & Solutions:

1) Answer (B)

Probability that at least 1 is blue => 1 – (probability that none is blue) => 1 – $(\frac{^{11}C_4}{^{15}C_4})$ = $\frac{69}{91}$

2) Answer (E)

Number of ways of selecting 2 red balls in 2 chances = $^{6}C_2$

Probability of 2 red balls in 2 chances = $\frac{^{6}C_2}{^{15}C_2}$ = $\frac{1}{7}$

3) Answer (C)

Number of ways of selecting 2 blue urns and 1 yellow urn = $^{4}C_2* ^{3}C_1$ = 18

Probability of selecting 2 blue urns and 1 yellow urn  = $\frac{18}{^{15}C_3}$ = $\frac{18}{455}$

4) Answer (A)

Number of ways of selecting 1 green, 2 blue and 1 red = $^2C_1*^4C_2*^6C_1 = 2*6*6 = 72$

Probability of selecting 1 green, 2 blue and 2 red = $\frac{72}{^{15}C_4}$ = $\frac{24}{455}$

5) Answer (D)

Both are green and or are yellow => $^2C_2 + ^3C_2 = 4$

Probability that both are green or both are yellow = $\frac{4}{^{15}C_2}$ = $\frac{4}{105}$

6) Answer (D)

Total number of letters = 6 of which 3 are the same.

Therefore, number of ways of arranging = $\frac{6!}{3!}$ = $\frac{720}{6}$ = 120

7) Answer (C)

None yellow => both blue => is possible in $^7C_2$ ways

=> probability = $\frac{^7C_2}{^{12}C_2}$ = $\frac{7}{22}$

8) Answer (D)

Number of ways in which the sum is  7 = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) => 6 ways

Total number of possibilities = 6 * 6 = 36

Probability that sum is 7 = $\frac{6}{36}$ = $\frac{1}{6}$

9) Answer (E)

Out of two red marbles two can be choosen in 2C2 ways
Out of two red marbles one can be choosen in 2C1 ways
Sum of blue, yellow and green marbles = 12
Out of 12 marbles one can be choosen in 12C1 ways
The probability that both are red or atleast one is red = [2C2 + (2C1 x 12C1)]/14C2 = $\frac{25}{91}$

10) Answer (B)

At least one marble is yellow which means that there can be one yellow marble or two yellow marbles or three yellow marbles.
We know that 0 < probability < 1.
Subtractiong the probability of having no yellow marble gives the probability of having at least on yellow marble.
1 – (11C3/14C3) = 1 – $\frac{165}{364}$ = $\frac{199}{364}$

11) Answer (C)

In order to have equal number of marbles in each colour one has to draw 2 marbles from each colour.
Out of four green marbles two can be choosen in 4C2 ways
Out of five green marbles two can be choosen in 5C2 ways
Out of two red marbles two can be choosen in 2C2 ways
Out of three yellow marbles 2 can be choosen in 3C2 ways
So the probability is = ( 4C2 + 5C2 + 2C2 + 3C2 )/(14C8) = $\frac{60}{1001}$

12) Answer (E)

Out of the fourteen marbles four are green in colour
Subtracting the green marbles from the total marbles we will be left with 10 marbles
Out of ten marbles three can be choosen in 10C3 ways
Probability of not choosing a choosing marble = 10C3/14C= $\frac{30}{91}$

13) Answer (A)

Out of five blue marbles two can be choosen in 5C2 ways
Out of two red marbles two can be choosen in 2Cways
So the probability of choosing two blue and two red marbles is = (5C2C2)/14C= $\frac{10}{1001}$

14) Answer (D)

Number of ways of arranging seven letters = 7!
Let us consider the two vowels as a group
Now the remaining five letters and the group of two vowels = 6
These six letters can be arranged in 6!2! ways( 2! is the number of ways the two vowels can be arranged among themselves)
The number of ways of arranging seven letters such that no two vowels come together
= Number of ways of arranging seven letters – Number of ways of arranging the letters with the two vowels being together
= 7! – (6!2!)
= 3600

15) Answer (D)

Probability of getting two white balls = 13P2
Probability of getting two black balls = 7P2
Total probability = (13P2 + 7P2)/20P2 = 198/380 = 99/190

16) Answer (A)

Circular permutation = n! (n – 1)!

∴ Number of ways = 6! (6 – 1)! = 6! × 5!

17) Answer (B)

Word name: ‘JOKE’ Vowels: O, E
Consonants: J, K
∴ Possible arrangement beginning with consonant: JOKE, KOJE, JEKO, KEJO = 4 Numbers
beginning with vowel: OJEK, OKEJ, EJOK, EKOJ = 4 Numbers
Required number = 4+4 = 8 numbers

18) Answer (B)

No two women would be seated next to each other if they sit between men. So, first the men can be arranged in 5! ways. There are six spots between the men. Number of ways to arrange the 3 women in those six spots is $ P_6^3 $ ways. So, total number of ways = 120*120 = 14400 ways.

19) Answer (E)

If there are n different letter in a word then n! different words can be formed.

In SIMPLE there are 6 different letters and hence 6! new words can be formed by different arangement.

Hence, 720 new words can be formed.

20) Answer (A)

If a word has n different letters in it, then they can be arranged in n! different ways.

Here SECOND has 6 different letters and they can be arranged in 6! ways i.e 720 different ways.

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