# TCS NQT Probability Questions Quiz 1

Question 1

Time: 00:00:00
2/3rd of the balls in a bag are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls are NOT defective, find the total number of balls in the bag given that the number of non-defective balls is 142.

216

216

422

422

432

432

644

644

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#### PrepInsta User

when u r saying that 5/9 of blue balls are defective then (1-5/9)4/9 balls of 2/3 balls are non defective . aswell as for pink . So i think that answer is incorrect

#### PrepInsta User

(4/9)X(2x/3)+(1/8)X(x/3)=146 (8x/9)+(x/8)=146 X 3 x=432

Question 2

Time: 00:00:00
100 students appeared for two different examinations 60 passed the first,50 the second and 30 both the examinations.Find the probability that a student selected at random failed in both the examination?

5/6

5/6

1/5

1/5

1/7

1/7

5/7

5/7

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Question 3

Time: 00:00:00
From a bag containing 8 green and 5 red balls, three are drawn one after the other .the probability of all three balls beings green if the balls drawn are replaced before the next ball pick and the balls drawn are not replaced, are respectively?

521/2197, 336/2197

521/2197, 336/2197

512/2197, 28/143

512/2197, 28/143

336/2197, 512/2197

336/2197, 512/2197

336/1716, 512/1716

336/1716, 512/1716

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Question 4

Time: 00:00:00
A bag contains 8 white balls and 3 blue balls. Another bag contains 7 white and 4 blue balls. What is the probability of getting blue ball?

3/7

3/7

7/22

7/22

7/25

7/25

7/15

7/15

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Question 5

Time: 00:00:00
There are two bags, one of which contains 5 red and 7 white balls and the other 3 red and 12 white balls. A ball is to be drawn from one or other of the two bags ; find the chances of drawing a red ball.

55/102

55/102

17/21

17/21

37/120

37/120

7/8

7/8

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Question 6

Time: 00:00:00
On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is?

2/5

2/5

4/5

4/5

5/2

5/2

6/5

6/5

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Question 7

Time: 00:00:00
3 dice are rolled. What is the probability that you will get the sum of the no’s as 10?

27/216

27/216

25/216

25/216

10/216

10/216

1/11

1/11

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Question 8

Time: 00:00:00
There are 5 letters and 5 addressed envelopes. If the letters are put at random in the envelops, the probability that all the letters may be placed in wrongly addressed envelopes is

65/120

65/120

44/120

44/120

59/120

59/120

40/120

40/120

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#### PrepInsta User

I think I have found a new and simple formula to suit this particular case. If there is one letter and one envelope then no way you can put it wrong(S1). If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2). If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3). If there are 4 then you can put them wrong in 9 ways(S4). If there are 5 then you can put them wrong in 44 ways(S5). If you observe you can find a pattern. S3=(S1+S2)*2 S4=(S2+S3)*3 S5=(S3+S4)*4 S6=(S4+S5)*5 In general, Sn=(Sn-2 + Sn-1)*(n-1) So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4=44

#### PrepInsta User

44 Number of ways in which \'n\' objects can be placed on \'n\' positions in such a manner that none of them is correct is given by the Dearrangement formula. Dearr(n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!) In this question, we need to place 5 objects (letters) in 5 positions (addresses) such that none of them is correct. This can be done in Dearr(5) = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!) => Dearr(5) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120) => Dearr(5) = 60 - 20 + 5 - 1 = 44 So, we can send the 5 letters, such that all are delivered at wrong adresses, in 44 different ways. Copied from somewhere.

Question 9

Time: 00:00:00
Thangam and Pandiyamma go for an interview for two vacancies. The probability for the selection of Thangam is 1/3 and whereas the probability for the selection of Pandiyamma is 1/5. What is the probability that none of them are selected?

3/5

3/5

7/12

7/12

8/15

8/15

1/5

1/5

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2/3+ 4/5 = 8/15

fghjk

8/15

#### PrepInsta User

C

Question 10

Time: 00:00:00
A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?

291/1100

291/1100

292/1100

292/1100

290/1100

290/1100

301/1100

301/1100

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#### PrepInsta User

C

["0","40","60","80","100"]
["Need more practice! \r\n","Keep trying! \r\n","Not bad! \r\n","Good work! \r\n","Perfect! \r\n"]

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