Latest TCS Aptitude Question SOLUTION: 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 at the digit 2 appearing on it?

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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 at the digit 2 appearing on it?

Asked In TCS RAvi (10 years ago)

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Karan  (10 years ago)

no of numbers having 2 in 1 to 100=19
no of numbers having 2 in 101 to 199=19
no of numbers having 2 in 200 to 299=100
no of numbers having 2 in 300 to 1100=19*8
so prob = (190+100)/1100= 290/1100 = 29/110

Is this solution Helpfull? Yes (63) | No (29)  

kalaivani  (10 years ago)

hi friend in 1 to 100 ,no of numbers having 2 will be 20 because in 20 series it will be of 11 twos becos in 22 ,two times no 2 is repeated.

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komarvallychaitanya  (10 years ago)

no of 2's in 1 to 99=20,lly in a 101 to 199,
200 to 299 =120,on sooon
total no.of 2 in 1 to 1100 =300
prob=300/1100=3/11

Is this solution Helpfull? Yes (6) | No (8)  

varun saxena  (8 years ago)

no of 2's from 1 to 100 is 20
no of 2's from 101 to 199 is 20
no of 2's from 200 to 300 is 120
no of 2's from 200 to 300 is 20
and so on
so total no of 2's b/w 1 to 1100 is 320
so probability 320/1100=16/55

Is this solution Helpfull? Yes (5) | No (4)  

anjali raj  (9 years ago)

no. of 2's in 1 to 1100 =300
so if 1 ticket is drawn at random then the probability will be 300/1100=3/11

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swati dwivedi  (9 years ago)

22,122,222,322.....922 are the single numbers. dont count 22, 2 times.
karan ans is correct.

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sunny kumar  (10 years ago)

probability=0.19
no's digits appearinin on it is 19 frm 1 to 100
then frm1,2,3......1100 =19*11=209
probability = 209/1100=0.19

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Aravinth  (7 years ago)

Ans is 32/110=16/55
1 to 999- digit wise counting
three digit:
so
i) 2 as first digit so possibilities are in second and third are 10 which is =1*10*10=100
ii) 2 as second digit so possibilities are in first and third are 10 which is =10*1*10=100
iii) 2 as third digit so possibilities are in first and second are 10 which is =10*10*1=100
then Now we consider 1000-1099
2 is not appear in first two digit
So i) 2 as third digit, possibilities of fourth digit is 10=1*1*1*10=10
i) 2 as forth digit, possibilities of third digit is 10=1*1*10*1=10

Finally total=100+100+100+10+10=320
Probability=320/1100=32/110=16/55

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