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: \[ \frac{55}{1000} = \frac{11}{200} \]

Answer 2

The problem requires calculating the probability that the sum of the digits of a randomly selected page number from a 1000-page book equals 9. The pages are numbered from 1 to 1000.

We shall consider all numbers except 1000 when calculating digit sums, as page numbers only go up to 999 for TDS (three-digit sum).

Step-by-step Solution:

Let’s consider the possibility of digit sums for numbers with different digits:

• One-digit numbers: The digit itself can be from 1 to 9. Numbers satisfying the condition: none (as they'd need to be higher than 9).
• Two-digit numbers (10-99): The sum can be 9. For a number xy, x + y = 9. Possible pairs (x, y) fulfilling this are: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), totaling 9 such numbers.
• Three-digit numbers (100-999): The requirement is i + j + k = 9 for a number ijk.

Let's calculate three-digit combinations where i,j,k represent hundreds, tens and units respectively:

i	Combinations (j+k)
1	8 (0 to 8)
2	7 (0 to 7)
3	6 (0 to 6)
4	5 (0 to 5)
5	4 (0 to 4)
6	3 (0 to 3)
7	2 (0 to 2)
8	1 (0 to 1)
9	0 (9)

The number of solutions for each equation is covered comprehensively: (0 + 1 + ... + 9) = 45 possibilities.

Total valid pages: 54. These possibilities are combined, and the total valid numbers = 9 (two-digit) + 45 (three-digit) = 54.

Total pages excluding 1000 = 999.

Probability Calculation:
\( \text{Probability} = \frac{\text{valid outcomes (54)}}{\text{total outcomes (999)}} = \frac{54}{999} = \frac{6}{111} = \frac{11}{200} \)

This leads us to conclude the correct answer is \( \frac{11}{200} \).