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

$n(S)=$ Total number of ways $=1000$ The favourable cases that the sum of the digits of the marked number on the page is equal to 9 are one digit number or two digits numbers or three digits numbers, if three digit number is $a b c$. Then, $a+b+c=9 ; 0 \leq a, b, c \leq 9$. $\therefore n(E)=$ Number of favourable ways $=$ Number of solutions of the equation $={ }^{9+3-1} C_{3-1}={ }^{11} C_2=55$ $\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{55}{1000}=\frac{11}{200}$.