a book contains1000pages. a page is chosen at random. find the probability that the sum of the digits of the marked numberon the pageis equal to 9.

Answer :

We know probability p ( E ) = Total number of desired events n( E)Total number of events n( S)
SO,
Here Total number of events n ( S ) = 1000
And
The page number whose sum is 9  , are

One digits numbers 0 to 9 = { 9 } =  1
And
Two digits numbers 10 to 99 = { 18 , 27 , 36 , 45 , 54 , 63 , 72 , 81 , 90 } = 9
And
Three digits numbers : From 100 to 200   = { 108 , 117 , 126 , 135 , 144 , 153 , 162 , 171 , 180 } = 9
And
From 200  to 300 = { 207 , 216 , 225 , 234 , 243 , 252 , 261 , 270 } = 8
And
From 300 to 400 = { 306 , 315 , 324 , 333 , 342 , 351 , 360 } = 7
And
from 400 to 500 = { 405 , 414 , 423 , 432 , 441 , 450 } = 6
And
From 500 to 600 = { 504 , 513 , 522 , 531 , 540 } = 5
And
From 600 to 700 = { 603 , 612 , 621 , 630 } = 4
And
From 700 to 800 = { 702 , 711 , 720 } = 3
And
From 800 to 900 = { 801 , 810 } = 2
And
From 900 to 999 = { 900 } = 1

So, Total number of desired events  n( E ) = 1  + 9 + 9 + 8 + 7 + 6 + 5 + 4+ 3+ 2+ 1 = 55
Then
Probability that the sum of the digits of the marked number on the pages equal to 9 = 551000 = 11200                                                    ( Ans )

  • 9
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