29 Jun 2017

Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Probability)

Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Probability):
Dear Readers, Important Practice Aptitude Questions for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.






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1. A card is drawn from a well shuffled pack of 52 cards. What is the probability of getting queen or club card?
17/52
15/52
4/13
3/13
None of these

1). Answer: c)
The probability of getting queen card = 4/52
The probability of getting club card = 13/52
The club card contains already a queen card, therefore required probability
is, 4/52 + 13/52 – 1/52 = 16/52 = 4/13

2.Two cards are drawn from pack of 52 cards. What is the probability that both are kings, when first drawn card is replaced?

1/169
3/13
3/676
4/676
None of these
2). Answer: a)
The probability of getting king card = 4/52
The first card is replaced so that, it doesn’t affect the second drawn card.
Hence, probability of getting 2nd king card = 4/52
∴ Required Probability = 4/52 × 4/52 = 1/169

3. A committee has 15 members, of whom only 5 woman. What is the probability that a committee of 11 members with at least 3 women is selected?
11/13
14/13
12/13
6/13
None of these
3).Answer: c)
There are 5 women and 10 men in the committee, out of these 11 members are selected. So, n(S) = 15C11
Probability of atleast 3 women in the committee is,
= (5C3 × 10C8)/ 15C11 + (5C4 × 10C7)/ 15C11 + (5C5 × 10C6)/ 15C11
= 1/15C11 × [5C2 × 10C2 + 5C1 × 10C3 + 5C5 × 10C4)]
= 1/1365 [450 + 600 + 210] = 1260/1365
= 12/13

4.An integer is chosen at random from the first fifty integers. What is the probability that the integer chosen is a prime or multiple of 4?
14/25
3/5
3/5
27/50
None of these
4). Answer: d)
There are 15 prime numbers in the first 50 integers i.e. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
There are 12 integers are multiples of 4 i.e 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44 an 48
∴ required probability= 15/50 + 12/50 = 27/50

Question (5-6): Three letters are written to three different persons and addresses on three envelop are also written without looking at the addresses.
5. What is the probability that all the letters go to right address?

1/3
2/3
1/6
1/9
None of these
Directions (Q. 5-6):

There are 6 combination of letters put into the envelop, n(s)= 6
5). Answer: c)
All the letters put into right envelop is only in E1.
Hence, required probability= 1/6

6.What is the probability that none of the letter goes to right address?
5/6
1/3
2/3
1/6
None of these
6. Answer: b)
None of the letters put into right envelop is E4 and E5
Hence, required probability= 2/6 = 1/3

7. Two friends A and B appear in an interview for two vacancies in the same post. The probability of A’s selection is 1/6 and that of B’s selection is 1/5. What is the probability that both of them will be selected and one of them will be selected?
11/30, 3/10
11/30, 2/3
1/30, 3/10
1/30, 2/3
None of these
7. Answer: c)
i) . The probability of both of them selected = 1/6 × 1/5 = 1/30
ii). The probability of A is not selected = 1-1/6 = 5/6
The probability of B is not selected = 1-1/5 = 4/5
∴ The probability of one of them selected = (1/6 × 4/5) + (1/5 × 5/6)
= 4/30 + 5/30 = 9/30 = 3/10

8. In a box containing ten toys, two are defective. What is the probability that among 5 toys chosen at random none is defective?
1/3
2/9
2/3
5/84
None of these
8. Answer: b)
The required probability = 8C5 / 10C5 = 8C3 / 10C5 = 2/9

9. A pot contains 5 white and 3 red balls while another pot contains 4 white and 6 red balls. One pot is chosen at random and a ball is drawn from it. If the ball is white, what is the probability that it is from the first pot?
5/16
41/80
25/41
1/5
None of these
9. Answer: c)
The probability of choosing one pot = 1/2
The probability of choosing white ball in first pot = 1/2 × [ 5C1 / 8C1] = 5/16
The probability of choosing white ball
= 1/2 × [5C1 / 8C1] + 1/2 × [4C1 / 10C1]
= 1/2 × [5/8] + 1/2 × [4/10]
= 5/16 + 1/5 = 41/80 ∴ The probability that it is from the first pot = (5/16) / (41/80)
= 5/16 × 80/41 = 25/41

10.Bag A contains 3 green and 7 blue balls. While bag B contains 10 green and 5 blue balls. If one ball is drawn from each bag, what is the probability that both are green?
29/30
1/5
1/3
1/30
None of these
10. Answer: b)
The required probability = 3C1/10C1 × 10C1/15C1
= 3/10 × 10/15 = 1/5





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