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 has the digit 2 appearing on it?

within 1 to 100 there are 19 numbers containing digit 2.2 first is 20 to 29 there are 10 numbers and 2 12,32,-,- ------92.
and 200 to 299 there are 100 numbers. so 1100 contain 290 numbers containing digit 2.(100---19 so 100 contain 190 ,excluding 200 to 299) 190+100=290.
so 290/1100 is answer.

ther are 19 times 2's appearing in (1 to 100, 301 to 400, 401 to 500, 501 to 600, 601 to 700, 701 to 800, 801 to 900, 901 to 1000 and 1001 to 1100 )
so there are 9*19= 171 times
there are 20 times 2's appearing from 101 to 200
and 99 times 2's appearing from 201 to 300

therefore there are 171+20+99 = 290
so C. 290/1100 is the correct answer

no. of numbers between 1 to 999 in which at-least one 2 is present =
[ fix 2 at 100th position and the unit and 10th places can be filled in 10*10 ways (as each place can be filled with 0 to 9 ) ]
+ [ fix 2 at 10th position and then 100th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at the 100th place) * 10 ( as unit place can still use values 0 to 9 ) ]
+ [ fix 2 at unit place and then 100th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at 100th place) * 10th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at 10th place) ]
=10*10+9*10+9*9
=100 + 90 + 81
=271
Now, no. of numbers between 1000 to 1099 in which at-least one 2 is present = 2 is fixed at 10th place and 10 ways of filling unit place
+ 9 ways of filling 10th place (0 to 9 excluding 2 ) and 2 is fixed at unit place
=10 + 9
=19

So, total no. of tickets on which digit 2 is appearing = 271 + 19 = 290
Therefore, required probability = 290/1100

All are saying that digit 2 is 19 times on 1to 100 how its 20times can u see 22 having two 2s
2 12 20 21 22 23 24 25 26 27 28 29 32 42 52 62 72 82 92 ???? please explain how u imagine this

(19*10+100)/1100

Numbers which dont have 2 from 1 to 9 = 8
Numbers which dont have 2 from 10 to 99:
Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbres.
Numbers which dont have 2 from 100 to 999 =_ _ _ = 8 × 9 × 9 = 648
Numbers which dont have 2 from 1000 to 1099 =10_ _ = 9 × 9 = 81
Finally 1100 does not have 2. So 1.
Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810
Tickets with 2 in them = 1100 - 810 = 290
Required probability = 290 / 1100

1to 100 2,12,20,21,22,23,24,25,26,27,28,29,32,42,52,62,72,82,92 total 19
like wise up to 1100 total 190 2s
in 200 series 100, 2s
total (190+100)/1100=290/1100

291/1100 guss

within 1 to 100 there are 20 numbers containing digit 2. 200 to 299 there are 100 numbers starting with digit 2. so total (20*10)+100=300. hence the answer will be 300/1100 that equals to 3/11.

2, 12 22,,32......92 in unit digit place ,20, 21,22...29 in 10th digit place, since 22 is two times one 22 ngltd total=19 from 200 to 299=100 ,since it is from 1 to 1100 , the total 2's in 10th digit is 190 (19*10)=19=290 ,the probability=(290/1100)

290/1100..

291/1100.

All the answers given for this question is wrong, everyone is copy pasting above answers....yes 1 to 100 every digit repeats itself 20 times except " 1"

First check for no appearance of 2
1 - 9 -> 8 ways
10 - 99 -> 8 * 9 ways -> 72 ways
100 - 999 ->8*9*9 ways -> 648 ways
1000 - 1099 ->1*1*9*9 ways ->81 ways
and 1100 ->1
Total = 810 ways of no appearence of 2
total ways = 1100

then , Probability = 1- 810/1100 = 290/1100

option C