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Step-by-Step Solution
Step 1: List the Sample Space
We have 50 tickets numbered from 00 to 49. Therefore, the sample space S consists of:
S = {00, 01, 02, …, 48, 49}, and hence the number of elements in S is
$n(S) = 50$.
Step 2: Identify the Event “Product of Digits is 0”
A ticket numbered $ab$ (where $a$ is the tens digit and $b$ is the units digit) has a product of digits
$a \times b = 0$ if at least one of its digits is 0. For the tickets 00 to 49, those with product 0 are:
00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40.
Hence,
$$
n(\text{Product = 0}) = 14.
$$
Step 3: Identify the Event “Sum of Digits is 8”
We want the sum of digits $a + b = 8$. For tickets in the range 00 to 49, the pairs $(a, b)$ that satisfy
this are:
08, 17, 26, 35, 44.
Therefore,
$$
n(\text{Sum = 8}) = 5.
$$
Step 4: Find the Intersection “Sum = 8 and Product = 0”
We check which ticket(s) among those with sum of digits 8 also have at least one digit 0. Out of the pairs
08, 17, 26, 35, 44, only 08 has a product of 0. So:
$$
n(\text{Sum = 8} \cap \text{Product = 0}) = 1 \quad \text{(the ticket 08)}.
$$
Step 5: Apply the Conditional Probability Formula
The conditional probability of the sum of digits being 8, given that the product of the digits is 0, is:
$$
P(\text{Sum = 8} \mid \text{Product = 0})
=
\frac{P(\text{Sum = 8} \cap \text{Product = 0})}{P(\text{Product = 0})}.
$$
We calculate each term as follows:
$
P(\text{Product = 0}) = \frac{n(\text{Product = 0})}{n(S)} = \frac{14}{50}.
$
$
P(\text{Sum = 8} \cap \text{Product = 0}) = \frac{n(\text{Sum = 8} \cap \text{Product = 0})}{n(S)}
= \frac{1}{50}.
$
Step 6: Compute the Final Answer
Substituting these values into the conditional probability formula:
$$
P(\text{Sum = 8} \mid \text{Product = 0})
=
\frac{\frac{1}{50}}{\frac{14}{50}}
= \frac{1}{14}.
$$
Hence, the required probability is $ \frac{1}{14} $.
Conclusion
Therefore, the correct answer is $ \frac{1}{14} $.
