One ticket is selected at random from 100 tickets numbered 00,01,02,...,98,99. If `x_1, a n dx_2` denotes the sum and product of the digits on the tickets, then `P(x_1=9//x_2=0)` is equal to `2//19` b. `19//100` c. `1//50` d. none of these

Explanation

To solve the problem, we need to find the probability \( P(x_1 = 9 \mid x_2 = 0) \), where \( x_1 \) is the sum of the digits of the ticket number and \( x_2 \) is the product of the digits.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have tickets numbered from 00 to 99. Each ticket can be represented as a two-digit number \( xy \), where \( x \) is the tens digit and \( y \) is the units digit. We need to find the probability that the sum of the digits \( x_1 = x + y = 9 \) given that the product of the digits \( x_2 = x \cdot y = 0 \).

**Hint**: Remember that the product of the digits is zero if at least one digit is zero.

2. **Finding \( P(x_1 = 9 \cap x_2 = 0) \)**:
- For \( x_2 = 0 \) (product is zero), either \( x = 0 \) or \( y = 0 \).
- If \( x = 0 \), then \( y \) can be 0 to 9. The valid pairs are: \( (0,0), (0,1), (0,2), (0,3), (0,4), (0,5), (0,6), (0,7), (0,8), (0,9) \) — **10 outcomes**.
- If \( y = 0 \), then \( x \) can be 1 to 9. The valid pairs are: \( (1,0), (2,0), (3,0), (4,0), (5,0), (6,0), (7,0), (8,0), (9,0) \) — **9 outcomes**.
- Total outcomes for \( x_2 = 0 \): \( 10 + 9 = 19 \).

**Hint**: Count the pairs carefully based on the conditions for \( x_2 = 0 \).

3. **Finding \( P(x_1 = 9 \cap x_2 = 0) \)**:
- We need pairs where \( x + y = 9 \) and \( x \cdot y = 0 \).
- The only pairs that satisfy both conditions are \( (0,9) \) and \( (9,0) \) — **2 outcomes**.

**Hint**: Check which combinations of \( x \) and \( y \) satisfy both equations.

4. **Calculating \( P(x_1 = 9 \cap x_2 = 0) \)**:
\[ P(x_1 = 9 \cap x_2 = 0) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{100} = \frac{1}{50} \]

**Hint**: Use the total number of tickets (100) for the denominator.

5. **Calculating \( P(x_2 = 0) \)**:
\[ P(x_2 = 0) = \frac{19}{100} \]

**Hint**: This is the total number of outcomes where the product of digits is zero divided by the total number of tickets.

6. **Finding \( P(x_1 = 9 \mid x_2 = 0) \)**:
Using the formula for conditional probability:
\[ P(x_1 = 9 \mid x_2 = 0) = \frac{P(x_1 = 9 \cap x_2 = 0)}{P(x_2 = 0)} = \frac{\frac{1}{50}}{\frac{19}{100}} = \frac{1}{50} \cdot \frac{100}{19} = \frac{2}{19} \]

**Hint**: Remember to multiply by the reciprocal of the denominator when dividing fractions.

### Final Answer:
The probability \( P(x_1 = 9 \mid x_2 = 0) \) is \( \frac{2}{19} \).