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 possible page selections: 1000. Favorable outcomes occur when the sum of the digits of the marked page number equals 9. This includes one-digit, two-digit, or three-digit numbers. For a three-digit number \(abc\), the condition is \( a + b + c = 9, \quad 0 \leq a, b, c \leq 9 \). The number of solutions for \(a + b + c = 9\) is given by \(11 C_2 = 55\). Therefore, the probability is: \[ \frac{55}{1000} = \frac{11}{200} \]