Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from June 2018. To enrich your preparation here we have providing new series of Probability – Quantitative Aptitude Questions. Candidates those who are appearing in SBI Clerk Prelims Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.
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1) A cartoon contains 7 red and 5 green apples. 3 apples are drawn at random. Find the probability that they are of the same colors.
2) 2 dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is divisible by either 3 or 5?
3) From a pack of 52 cards one card is drawn at random. What is the probability that the card drawn is a six or a diamond?
4) 2 cards are drawn together at random from a pack of 52 cards. What is the probability of both the cards being jack?
5) In a bouquet, there are 5 red, 3 white and 7 orange roses. One rose is picked up randomly. What is the probability that is neither red nor orange?
6) In a bag contains 6 black toys, 7 violet toys and 5 blue toys. Three toys are drawn at random from the bag. The probability that all of them are blue is?
7) A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a heart nor a king?
8) A box contains 7 black and 5 white hair clips. 3 hair clips are drawn at random. What is the probability that one is black and the other 2 are white?
9) A pouch contains 4 black, 2 red and 5 blue pens. 2 pens are drawn at random. What is the probability that none of the pens drawn is blue?
10) A box contains 15 tube lights, out of which 6 are repair. 3 tube lights are chosen at random from this box. Find the probability that at least one of these is repair?
Answers:
1) Answer: B
Let s be the sample space.
Then n(S) = no. of ways of drawing 3 apples out of 12
= > 12C3 = 12*11*10 / 3*2*1
= > 220
Let E = event of getting both apples of the same color. Then
n(E) = no. of ways of drawing 3 apples out of 7 or 3 apples out of 5
=7C3 +5C3
=7*6*5 / 3*2*1 + (5*4*3 / 3*2*1)
=35 + 10
=45
P(E) = n(E) / n(S)
= 45 / 220
= 9/44
2) Answer: B
Clearly n(s) = 6*6 = 36
Let E be the event that the sum of the numbers on the 2 faces is divisible by either 3 or 5. Then
E = {(1,2), (1,4), (1,5), (2,1), (2,3), (2,4), (3,2), (3,3), (3,6), (4,1), (4,2), (4,5), (4,6), (5,1), (5,4), (5,5), (6,3), (6,4), (6,6)}
n(E) = 19
Hence P(E) = n(E) / n(s)
= 19/ 36
3) Answer: B
Here n(S) = 52
There are 13 diamond cards (including one six) and also 3 more sixes are there.
Let E = event of getting a six or a diamond
Then n(E) = 13 +3
n(E) = 16
Therefore P(E) = n(E) / n(S)
=16 / 52
P(E) = 4/13
4) Answer: D
Let s be the sample space
Then n (S) = 52C2
=52*51/2*1
=2652/2
=1326
Let E = event of getting 2 jack cards out of 4
n(E) = 4C2 = 4*3 / 2*1
=12/2
=6
P(E) = n(E)/n(S) = 6/1326
=1/221
5) Answer: B
Total no. of roses = 5+3+7
=15
Let E = event that the rose drawn is neither red nor orange
n(E) = 3
Therefore P(E) = n(E) / n(s)
=3/15
=1/5
The probability that is neither red nor orange is 1/5
6) Answer: B
Let s be the sample space
Then n(S) = no. of ways of drawing 3 toys out of 18
=18C3
=18*17*16 / 3*2*1
=816
let E = event of getting all the 3 blue balls
n(E) = 5C3
=5*4*3 / 3*2*1
=10
P(E) = n(E) / n(S)
=10/816
=5/408
7) Answer: C
There are 13 heart and 3 king
Probability of getting heart or a king:
= (13+3)/52
=16/52
=4/13
So probability of getting neither hearts nor a king:
=1- 4/13
=9/13
8) Answer: C
Let s be the sample space. Then
n(S) = no. of ways of drawing 3 hair clips out of 12
=12C3 = 12*11*10 / 3*2*1
=220
Let E = event of drawing 1 black and 2 white hairclips
n(E) = no. of ways of drawing 1 black out of 7 and 2 white out of 5
=7C1 and 5C2
=7*(5*4 / 2*1)
=7*10
=70
P(E ) = n(E) / n(S)
= 70/220
= 7/22
9) Answer: C
Total no. of pens = 4+ 2+5
=11
Let s be the sample space. Then
n(S) = no. of ways of drawing 2 pens out of 11
=11C2 = 11*10 / 2*1
=55
Let E = event of drawing 2 pens, none of which is blue
n(E) = no. of ways of drawing 2 pens out of 6 pens
=6C2
=6*5 / 2*1
=15
P(E) = n(E) / n(S)
=15 / 55
= 3/11
10) Answer: A
P(none is repair) = 9C3 / 15C3
= (9*8*7 / 3*2*1) / (15*14*13 / 3*2*1)
= (504/6) / (2730 / 6)
= 504/6 * 6/2730
= 504/2730
= 12/65
P(at least one is repair) = 1 – 12/65
= 53/65
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This post was last modified on April 20, 2018 6:06 pm