One ticket is selected at ransom form 50 tickets numbered `00,01,02,…,49.` Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, is

Options

A. `1/14`
B. `1/7`
C. `5/14`
D. `1/50`

Explanation

To solve the problem, we need to find the probability that the sum of the digits on a randomly selected ticket is 8, given that the product of the digits is 0. We will denote the events as follows:

- Let \( A \) be the event that the sum of the digits is 8.
- Let \( B \) be the event that the product of the digits is 0.

### Step 1: Identify the total number of tickets
The tickets are numbered from 00 to 49, which gives us a total of 50 tickets.

### Step 2: Determine the condition \( B \) (product of digits is 0)
The product of the digits is 0 if at least one of the digits is 0. We will list the tickets where the product of the digits is 0:

- Tickets with the first digit as 0: 00, 01, 02, 03, 04, 05, 06, 07, 08, 09 (10 tickets)
- Tickets with the second digit as 0: 10, 20, 30, 40 (4 tickets)
- The ticket 00 has already been counted in the first group.

Thus, the tickets where the product of the digits is 0 are:
- 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40

Counting these gives us a total of 14 tickets where the product of the digits is 0. Therefore, \( n(B) = 14 \).

### Step 3: Determine the condition \( A \) (sum of digits is 8)
Next, we find the tickets where the sum of the digits is 8. The valid combinations of digits (tens and units) that sum to 8 are:
- 08 (0 + 8)
- 17 (1 + 7)
- 26 (2 + 6)
- 35 (3 + 5)
- 44 (4 + 4)

Now we check which of these tickets also satisfy the condition \( B \) (product of digits is 0):
- 08 (product = 0)
- 17 (product ≠ 0)
- 26 (product ≠ 0)
- 35 (product ≠ 0)
- 44 (product ≠ 0)

Only the ticket 08 satisfies both conditions \( A \) and \( B \). Therefore, \( n(A \cap B) = 1 \).

### Step 4: Calculate the conditional probability
The conditional probability \( P(A|B) \) is given by the formula:
\[ P(A|B) = \frac{n(A \cap B)}{n(B)} \]
Substituting the values we found:
\[ P(A|B) = \frac{1}{14} \]

### Final Answer
Thus, the probability that the sum of the digits on the selected ticket is 8, given that the product of the digits is 0, is \( \frac{1}{14} \).

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