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 290/1100 d 301/1100
Answers
i hope answer is write because according to me there are no options proper
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
hope it helped you