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Q.
A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of
the marked number on the page is equal to 9 is :
Probability - Part 2
Solution:
$ n ( S )=1000 ;\{000,001, \ldots \ldots . .999\}$
$n ( A )= \text { coeff of } x ^9 \text { in }\left(1+ x + x ^2+\ldots .+ x ^9\right)^3 $
$= \text { coeff of } x ^9 \text { in }\left(1- x ^{10}\right)^3(1- x )^{-3} $
$= { }^{11} C _9=55 $
$P ( E )= \frac{55}{1000}=\frac{11}{200} $