TCS Probability Questions with Answers Quiz-1

432

Answer 432

it is 216

B

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

ans is 221

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

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

5/7

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)

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

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

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).

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%

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

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

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

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