A bag contains 1100 tickets. tickets are numbered from 1, 2 ... 1100 and one card is drawn randomly what is the probability of having 2 as a digit? (a) 190 (b) 200 (c) 250 (d) 400
Answers
Total number of tickets in the bag n(s) = 1100.
Let A be the event of drawing a card of having 2 as a digit.
(1) Number of times 2 appears in 1 - 100 = 19(20-29,2,12 32,42,52,62,72,82,92)
(2) Number of times 2 appears in 101 - 200 = 20.
(3) Number of times 2 appears in 201 - 300 = 99
(4) Number of times 2 appears in 301 - 400 = 19
(5) Number of times 2 appears in 401 - 500 = 19
(6) Number of times 2 appears in 501 - 600 = 19
(7) Number of times 2 appears in 601 - 700 = 19
(8) Number of times 2 appears in 701 - 800 = 19
(9) Number of times 2 appears in 801 - 900 = 19
(10) Number of times 2 appears in 901 - 1000 = 19
(11) Number of times 2 appears in 1000 - 1100 = 19.
So, required number of ways n(A) :
= > 20 + 99 + 9(19)
= > 290.
Therefore, the required probability P(A) = n(A)/n(S)
= .290/1100.
Your options are incorrect!
Hope this helps!
I think there is some problem with the options given correct options are as follows:
A. 291/1100
B. 292/1100
C. 290/1100
D. 301/1100
The correct answer is option C. 290/1100
For every 100 19 digits have 2.
Because we r counting the number of digits having, not the number of 2's in 100(here 22,122 these digits will be counted as 1 not 2)
So the answer is ((10*19)+100)/1100.as 200 to 299 there are 100 digits having 2.
That's why its 10*19+100
Solution 2:
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.