One ticket is selected at random from 100 tickets numbered 00, 01, 02, 03,…………..99. Suppose x is the sum of the digits and y is the product of the digits, then the probability that x = 9 and y = 0 is

Options

A. `1/99`
B. `1/50`
C. `3/50`
D. `1/25`

Answer

To solve the problem, we need to find the probability that the sum of the digits \( x \) is equal to 9 and the product of the digits \( y \) is equal to 0 for tickets numbered from 00 to 99.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let the digits of the ticket be represented as \( a \) and \( b \), where the ticket number is \( ab \).
- The sum of the digits is given by \( x = a + b \).
- The product of the digits is given by \( y = a \times b \).

2. **Conditions for \( x = 9 \)**:
- We need to find pairs \( (a, b) \) such that \( a + b = 9 \).
- The possible pairs are:
- \( (0, 9) \)
- \( (1, 8) \)
- \( (2, 7) \)
- \( (3, 6) \)
- \( (4, 5) \)
- \( (5, 4) \)
- \( (6, 3) \)
- \( (7, 2) \)
- \( (8, 1) \)
- \( (9, 0) \)

3. **Conditions for \( y = 0 \)**:
- The product \( y = 0 \) if either \( a = 0 \) or \( b = 0 \).
- Therefore, we need to find pairs from the previous step that include at least one zero.

4. **Finding Valid Pairs**:
- From the pairs listed for \( x = 9 \):
- \( (0, 9) \) → \( y = 0 \)
- \( (9, 0) \) → \( y = 0 \)
- The valid pairs that satisfy both conditions \( x = 9 \) and \( y = 0 \) are:
- \( (0, 9) \)
- \( (9, 0) \)

5. **Counting Total Outcomes**:
- The total number of tickets is 100 (from 00 to 99).

6. **Counting Successful Outcomes**:
- The successful outcomes where \( x = 9 \) and \( y = 0 \) are 2: \( (0, 9) \) and \( (9, 0) \).

7. **Calculating the Probability**:
- The probability \( P \) is given by:
\[ P(x = 9 \text{ and } y = 0) = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{2}{100} = \frac{1}{50} \]

### Final Answer:
The probability that \( x = 9 \) and \( y = 0 \) is \( \frac{1}{50} \).