Question

A book contains $1000$ pages numbered consecutively. The probability that the sum of the digits of the marked number of a page is $9$ is

• A. $\frac{55}{1000}$
• B. $0$
• C. $\frac{33}{1000}$
• D. $\frac{44}{1000}$

Answer 1

The correct option is A

$\frac{55}{1000}$

Explanation for the correct option:

Step 1: Find The probability that the sum of the digits of the marked number of a page is $9$

Given:

Total number of pages in a book, $n\left(S\right)=1000$

The event is getting the sum of digits is $9$

The possible outcomes are

One digit number $=9$

Number of one digit page $=1$

Two-digit number $=18,27,36,45,54,63,72,81,90$

Number of two-digit pages $={}^{9}C_1$

$=9$

Step 2: Find the required probability:

Three-digit numbers $$=9+8+7+6+5+4+3+2+1$$

$=\frac{n(n+1)}{2}$

$=\frac{9\times 10}{2}$

$=45$

Number of three-digit pages $=45$

Step 3. Total number of pages whose sum of digits is $9$ are

$\begin{array}{rcl}n\left(A\right)& =& 1+9+45\\ & =& 55\end{array}$

Thus the required probability $\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{55}}{\mathbf{1000}}$ $\left[\mathbf{\because }\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\right]$

Hence, option (A) is the correct answer.