Question

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

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

Hint:

To find probabilities in cases involving digit sums, break down the sum into its digit components and solve accordingly.

Answer 1

The Correct Option is B

Total number of ways to choose a page = 1000 The favorable cases that the sum of the digits of the marked number on the page is equal to 9 are one digit numbers or two-digit numbers or three-digit numbers, if the three-digit number is \(abc\). \[ a + b + c = 9, \quad 0 \leq a, b, c \leq 9 \] Hence, the number of favorable solutions for the equation \(a + b + c = 9\) is \(11 C_2 = 55\). Thus, the required probability is: \[ \frac{55}{1000} = \frac{11}{200} \]