Tickets are numbered from 1,2....1100 and one card is drawn randomly what is the probability of having 2 as a digit?

• for every 100 19 digits have 2...bcoz we r counting the nmbr of digits hving,not the nmbr of 2's in 100(here 22,122 these digits will be counted as 1 not 2)..so answr is ((10*19)+100)/1100.as 200 to 299 there are 100 digits having 2.dats why its 10*19+100
• 10 years agoHelpfull: Yes(59) No(12)
• for every 100 20 digits contain 2 as a digit.
  and from 200 to 299 also have 100 2 digits.
  here we have 11 hundreds so P(2)=(20*11)+100/(1100)=32/110.Ans
  
• 10 years agoHelpfull: Yes(36) No(27)
• for every 100 probability of having 2 as digit is 19 and 2 is used 20 times for every 100.according to qstn, there r 10 100's having 19 digits as 2 n 200 to 299 has 100 digits having 2. hence,
  ((10*19)+100)/1100=29/110
• 10 years agoHelpfull: Yes(15) No(6)
• the ans will be :: 290/1100.
  
  here we are looking for the digits(unique) having 2, not the no of 2's. so it not be the 320. we need to subtract those digits which have more than one 2's in a single digit like 22,222...!!
  
  so it can be calculate as = 19*10+100=290.
  prob=290/1100.
• 10 years agoHelpfull: Yes(10) No(4)
• As there are 19 -2's for every 100 we have(0-100,100-200,300-400,400-500,500-600,600-700,700-800,800-900,900-1000,1000-1100) there are 10-100's and 100-2's in (200-300) in total ((19*10)+100)/1100=290/1100.
• 6 years agoHelpfull: Yes(4) No(1)
• for every 100 20 digits contain 2 as a digit.
  here we have 11 hundreds so P(2)=(20*11)/(1100)=1/5
• 10 years agoHelpfull: Yes(2) No(11)
• frm 1 to 100 digit 2 comes 20 times, frm 101 to 200 also 2 comes 20 times bt frm 201 to 300 digit 2 comes more than 20 times thn how 20*11 is possible....
• 10 years agoHelpfull: Yes(2) No(11)
• (20*11+100)/1100=32/110
• 10 years agoHelpfull: Yes(2) No(5)
• answer is 29/110
• 7 years agoHelpfull: Yes(2) No(1)
• No. of 2's in 1-100=19 (count 22 only once)
  no.of 2's in 101-199=19(count 122 only once)
  no.of 2's in 200-299=100(no need to add the 2's which are in 2nd and 3 rd digits because they have only asked no of digits not no of 2's)
  there fore only 200-299 series has 200 2's.There is 11 100's in 1100 in which 200-299 has 100 2's.other 10 100's has 20 2's.
  so 10(20)+100=300
  
• 7 years agoHelpfull: Yes(2) No(3)
• first 1 to 100 is having 19 digits.(count of 2's).Then remaining 100 is there(between 1100)
  so, 19*10(10 -100 is there)+100=190
  190/1100=19/110
• 5 years agoHelpfull: Yes(1) No(0)
• for every 100 20 digits contain 2 as a digit.here we have 11 hundreds so we have 220 no 2 s...but we have not calculated another 110 2 s in series 200 to 299..so total no of 2 =220+110=330...P=330/1100=33/110 ans
• 10 years agoHelpfull: Yes(0) No(8)
• no of 2 between 1-1000=300 and 1000-1100=20 so total 2 btn 1-1100=320
  so probability=320/1100=32/110
• 10 years agoHelpfull: Yes(0) No(3)
• ans is 283/1100 as btw 1-100 2 occurs 19 times,100-200 19,200-299 100,18 times *9=283
• 10 years agoHelpfull: Yes(0) No(2)
• 1+19+280+20=320
  P=320/1100
• 5 years agoHelpfull: Yes(0) No(0)
• for every 100 10 digits contains 2's in its units place ,so for 1100 is 11*10=110
  for every 100 10 digits contains 2's in its 10's place ,so for 1100 is 11*10=110
  and in 100's place there are 100 2's in 1100
  by adding total number of 2' is=110+110+100=320
  probability is 320/1100
  we can write it as 32/110
• 5 years agoHelpfull: Yes(0) No(0)
• 99/1100
  the no of 2 digit cards is 99
• 5 years agoHelpfull: Yes(0) No(0)