TCS Probability Questions with Answers Quiz-1

Question 1

Time: 00:00:00
Solve the below given question :

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 defective, find the total number of balls in the bag given that the number of non-defective balls is 146.

216

216

432

432

649

649

578

578

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432

it is 216

B

ans is 221

Answer 432

432

Let total no of balls =x blue=2x/3 pink=x/3 Total no of defective balls = 10x/27 +7x/24 =143x/216 Non defective balls=x-143x/216=146 => x=432

a

Let the total numbers of the ball in a bag be x. Blue balls = 2/3 of x Pink ball = x - 2x/3 = x/3 Now, Defective Blue balls = 5/9 of (2/3 of x) = 5/9 × 2/3 × x = 10/27 × x Defective Pink balls = 7/8 of x/3 = 7x/24 Now, According to the Question, Defective Blue and Pink balls = 146 10x/27 + 7x/24 = 146 ? (80x + 63x)/216 = 146 ? 143x = 146 × 216 ? x = 220.53 ? 221 balls. Read more on Brainly.in - https://brainly.in/question/4976669#readmore

ans is 221

216

432

let the total balls be x 2/3x=blue & 1/3x=pink non defective balls=146 total no. of balls=defective+non defective hence defective=total-non defective(146) (2/3x)*5/9+(1/3x)*7/8=t-146 0.66203x=t-146 146=0.3303x x=432

let the total balls be x 2/3x=blue & 1/3x=pink non defective balls=146 total no. of balls=defective+non defective hence defective=total-non defective(146) (2/3x)*5/9+(1/3x)*7/8=t-146 0.66203x=t-146 146=0.3303x x=432

432

how it came 1/3 rd their

i think it is 216

432

defective blue ball + defective pink ball +non defective balls = total number of balls (5/9)(2x/3)+(7/8)(1x/3) + 146 = x 10x/27+7x/24 -x = -146 x=432 answer is b

let totalball= x non defective= 146 defective= x-146 blue ball= 2x/3 pink = x/3 defective blue ball = 5/9 x 2x/3= 10x/27 defective pink ball= 7/8 x x/3= 7x/24 > 10x/27 + 7x/24 = x-146 no of defective balls\\ 143x/216 = x - 146 143x = 216(x - 146) 216x-143x= 146 x 216 73x = 146 x 216 x= 432 (total no of balls)

Let 2/3x=blue 1/3y=pink Non defective=146 Total=non defective+defective Now we find defective balls Defective= total-non defective (146)*(2/3x)*(5/9)+(1/3)*(7/8)=total-(146) X=432 its defective Total=146+432 T=578

Let 2/3x=blue 1/3y=pink Non defective=146 Total=non defective+defective Now we find defective balls Defective= total-non defective (146)*(2/3x)*(5/9)+(1/3)*(7/8)=total-(146) X=432 its defective Total=146+432 T=578

b

(2/3)*(5/9)+(1/3)*(7/8)=(x-146) (10/27)+(7/24)=(x-146) X=578

Defective blue ball + Defective pink ball =(x-146) (5/9)(2x/3)+(7/8)(1x/3)=(x-146) 10x/27+7x/24=(x-146) x=221

432

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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?

0.8

0.8

0.2

0.2

0.6

0.6

0.7

0.7

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students atleast passed in all exams=n(aub)=60+50-30=80 remaining 20 are failed in both exams. so probability of student failed in both exams when a student selected randomly from them is 20/100=1/5.

T1=60-30=30 T2=50-30=20 Total pass=30+30+20=80 So P(s)=20÷100 1/5

p(1st) = 60 p(2nd) = 50 p(1st and 2nd)=30 so,no of std passed at least one subject is.... p(1st or 2nd)=60+50-30=80 so, no of std failed at least one subj is.....100-80=20 therefore, a std is selected 20/100 = 1/5 (ans)

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

336/2197, 512/2197

336/2197, 512/2197

512/2197, 336/1716

512/2197, 336/1716

336/1716, 512/1716

336/1716, 512/1716

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not replaced = (8/13)* (7/12)*(6/11)= 336/1716 with replace = (8/13)^3 = 512/2197

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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?

7/22

7/22

3/7

3/7

7/25

7/25

7/15

7/15

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Bag A contain 8white ball and 3blue ball total=11 Bag B contain 7white ball and 4blue ball total=11 Total=22 Total blue ball=7

total number of balls in two bags(white balls + blue balls)=22 only blue balls in both bags are 7 so out of 22 attempts in picking blue ball in both the bags, you get blue ball 7 times so the probability is 7/22

getting of blue ball is (total blue ball)/(total ball) = 7/22

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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.

37/120

37/120

55/102

55/102

17/21

17/21

7/8

7/8

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P(Getting a red ball from the first bag): (1/2)*(5/12) P(Getting a red ball from the 2nd bag): (1/2)*(3/15) P(Drawing a red ball): (1/2)*(5/12)+(1/2)*(3/15)= 37/120 (Ans).

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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?

50

50

45

45

40

40

60

60

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total probabilities for getting 5 = 4/36 total probabilities for getting 7 = 6/36 Total Probability = 10/36 We need only 5, hence prob of getting only 5 is (4/36)/(10/36) =40%

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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?

25/216

25/216

27/216

27/216

10/216

10/216

1/11

1/11

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total events = 6*6*6=216 possible cases for sum equal to 10 are (1,3,6)-6 combinations (3!) (1,4,5)-6 combinations (2,3,5)-6 combinations (2,4,4)-3 combinations (3!/2! as repetition) (3,3,4)-3 combinations (2,2,6)-3 combinations so total combinations are 27 so probability will be 27/216

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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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If there are 5 letters and 5 envelope then the sample space is 5*5=25and wrongevents =20 . So p(e)= 20/25 =4/5

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.

This group of questions are called \"derangement\" No. of possible arrangements so that out of n objects none goes to its correct/original position is D(n) D(n)= n! (1 - 1/1! + 1/2! - 1/3! + ... +(-1)n 1/n! ) D(5) = 5! (1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120) = 120(11/30) = 44

5!-1=119

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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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Probability of thangam being selected= 1/3 and him not being selected = 1-1/3=2/3 Probability of pandiyamma being selected=1/5 and him not being selected= 1-1/5=4/5 Probability that both are not selected = 2/3*4/5=8/15

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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?

290/1100

290/1100

291/1100

291/1100

292/1100

292/1100

301/1100

301/1100

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Tickets which contains 2 in between 0-100= 19 like wise; Tickets which contains 2 in between 200-299= 100 tickets from 0-1100 which contains 2 is = 19*10=190 1*100=100 so, 190+100=290 probability of getting tickets drawn from 0-1100 = 290/1100

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