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?

A. 291/1100
B. 292/1100
C. 290/1100
D. 301/1100

• 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.
  
• 8 years agoHelpfull: Yes(14) No(4)
• 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
• 8 years agoHelpfull: Yes(14) No(0)
• 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
  
• 8 years agoHelpfull: Yes(9) No(0)
• 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
• 7 years agoHelpfull: Yes(6) No(0)
• (19*10+100)/1100
  
• 8 years agoHelpfull: Yes(4) No(1)
• 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
• 8 years agoHelpfull: Yes(3) No(1)
• 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
  
• 8 years agoHelpfull: Yes(1) No(1)
• 291/1100 guss
• 6 years agoHelpfull: Yes(1) No(0)
• 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.
• 8 years agoHelpfull: Yes(0) No(5)
• 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)
• 8 years agoHelpfull: Yes(0) No(1)
• 290/1100..
  
• 8 years agoHelpfull: Yes(0) No(0)
• 291/1100.
  
• 8 years agoHelpfull: Yes(0) No(0)
• 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"
• 7 years agoHelpfull: Yes(0) No(0)
• 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
• 5 years agoHelpfull: Yes(0) No(0)
• 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
• 1 year agoHelpfull: Yes(0) No(0)