SBI PO Probability Questions and Answers PDF

Probability questions for Bank PO & Clerk exams of SBI PO based on asked previous questions in SBI PO & other banking exams. Download Probability questions and answers PDF with solutions and explanations (Important aptitude questions useful for prelims & mains)

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Question 1: 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

Question 2: 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 3: 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 4: 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

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Question 5: 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 6: 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 7: 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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Question 8: In how many different ways can the letters of the word ‘VENTURE’ be arranged?
a) 840
b) 5040
c) 1260
d) 2520
e)None of these

Question 9: In how many different ways can the letters of the word BAKERY be arranged ?
a) 2400
b) 2005
c) 720
d) 5040
e)None of these

Question 10: In how many different ways can the letters of the word DAILY be arranged ?
a) 60
b) 48
c) 160
d) 120
e)None of these

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Answers & Solutions
1 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 = 636\frac{6}{36}366​ = 16\frac{1}{6}61​ ​

2 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

3 Answer(A)
Circular permutation = n! (n – 1)!
∴ Number of ways = 6! (6 – 1)! = 6! × 5!

4 Answer(B)
Solution: 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

5 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 P63 P_6^3 P63​ ways. So, total number of ways = 120*120 = 14400 ways.

6 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.

7 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.

8 Answer(D)
In “VENTURE”, there are 7 letters of which E is repeated twice.
Number of ways of arranging the letters of the word VENTURE = 7!/2! = 5040/2 = 2520

9 Answer(C)
As there are 6 different letters in the word “Bakery” hence, total number of ways of arranging it will be = 6! = 720

10Answer(D)
If a word has n different letters then they can be arranged in n! different ways.
DAILY has 5 different letters which can bve arranged in 5! different ways.
5!=5*4*3*2*1=120
Therefore, option D is answer.

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