Question

A book contains $1,000$ pages. A page is chosen at random. The probabilities that the sum of the digits of the marked number on the page is equal to $9$ is

• A. $\frac{23}{500}$
• B. $\frac{11}{200}$
• C. $\frac{7}{100}$
• D. None of these

Answer 1

Answer: (B)

$\frac{11}{200}$

The number of number whose sum is $9$ is

One-digit number $=1$

Two-digit numbers $=9$

Three digits numbers $=9+8+7+...+1=\frac{9\times 10}{2}=45$

$\therefore$ Required probability $=\frac{55}{1000}=\frac{11}{200}$