Key Concepts and Formulas

• Conditional Probability: The probability of an event A occurring given that another event B has already occurred is denoted as $P(A|B)$. It is fundamentally about narrowing down the sample space to only those outcomes where event B has happened.
• Formula for Conditional Probability: For any two events A and B, the conditional probability is given by: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ Where $P(A \cap B)$ is the probability of both A and B occurring, and $P(B)$ is the probability of B occurring.
• Counting Principle for Equally Likely Outcomes: When all outcomes in the sample space are equally likely (as in selecting a ticket at random), the probability of an event can be calculated as the ratio of favorable outcomes to the total possible outcomes. For conditional probability, this translates to: $P(A|B) = \frac{\text{Number of outcomes in } (A \cap B)}{\text{Number of outcomes in } B} = \frac{n(A \cap B)}{n(B)}$ This formula effectively treats the outcomes of Event B as the new, reduced sample space.

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Step-by-Step Solution

Step 1: Understand the Problem and Define Events The problem asks for a conditional probability: "the probability that the sum of the digits on the selected ticket is $8$, given that the product of these digits is zero."

Let's clearly define the events:

• Event A: The sum of the digits on the selected ticket is $8$.
• Event B: The product of the digits on the selected ticket is zero.

Our goal is to find $P(A|B)$, which is the probability of Event A occurring given that Event B has already occurred. We will use the counting principle: $P(A|B) = \frac{n(A \cap B)}{n(B)}$.

Step 2: Define the Sample Space ($S$) The tickets are numbered from $00, 01, 02, \dots, 49$. This set of $50$ tickets represents our complete sample space, $S$. The total number of tickets in the sample space is $n(S) = 50$. Each ticket has a tens digit ($T$) and a units digit ($U$). For example, ticket $23$ has $T=2, U=3$. Ticket $07$ has $T=0, U=7$. Why this step is important: Clearly defining the full range of possibilities is crucial. It ensures we only consider valid tickets when listing outcomes for our events, preventing errors like including a ticket like $80$ if it's not in the sample space.

Step 3: Identify Outcomes for Event B (Product of Digits is Zero) and Calculate $n(B)$ Event B occurs if the product of the digits ($T \times U$) is zero. This happens if and only if either the tens digit ($T$) is zero, or the units digit ($U$) is zero (or both). We will list all tickets from our sample space ($00$ to $49$) that satisfy this condition.

1. Tens digit ($T$) is zero: These tickets are $00, 01, 02, 03, 04, 05, 06, 07, 08, 09$. There are $10$ such tickets.

2. Units digit ($U$) is zero (and tens digit $T$ is non-zero): We need to consider tickets where $U=0$ but $T \neq 0$ to avoid double-counting the ticket $00$, which was already included in the first category. These tickets are $10, 20, 30, 40$. There are $4$ such tickets.

Combining these two sets, the set of tickets for Event B is: $B = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$

The total number of outcomes in Event B is $n(B) = 10 + 4 = 14$. Why this step is important: $n(B)$ forms the denominator of our conditional probability. It represents the size of our reduced sample space—all the outcomes that are possible given that Event B has occurred.

Step 4: Identify Outcomes for Event $A \cap B$ (Sum of Digits is 8 AND Product of Digits is Zero) and Calculate $n(A \cap B)$ Now we need to find the tickets that satisfy both conditions simultaneously:

1. The sum of the digits is $8$ ($T+U=8$).
2. The product of the digits is zero ($T \times U = 0$).

From Step 3, we know that for the product of digits to be zero, one of the digits must be zero. Let's apply this to the sum condition:

• Case 1: Tens digit ($T$) is zero. If $T=0$, then for the sum $T+U=8$, the units digit $U$ must be $8$. This gives us the ticket $08$. This ticket is within our sample space ($00-49$) and is also in Event B (as its product of digits is $0 \times 8 = 0$). So, $08$ is an outcome in $A \cap B$.

• Case 2: Units digit ($U$) is zero. If $U=0$, then for the sum $T+U=8$, the tens digit $T$ must be $8$. This would imply a ticket like $80$. However, our sample space only includes tickets up to $49$. Therefore, $80$ is not a valid ticket in this problem.

Thus, the only ticket that satisfies both conditions (sum of digits is $8$ AND product of digits is zero) is $08$.

So, the set of outcomes for Event $A \cap B$ is: $A \cap B = \{08\}$

The total number of outcomes in Event $A \cap B$ is $n(A \cap B) = 1$. Why this step is important: $n(A \cap B)$ forms the numerator of our conditional probability. It represents the number of favorable outcomes within the reduced sample space defined by Event B.

Step 5: Calculate the Conditional Probability $P(A|B)$ Now, we apply the conditional probability formula using the counts of outcomes we determined: $P(A|B) = \frac{n(A \cap B)}{n(B)}$

Substitute the values we found:

• $n(A \cap B) = 1$
• $n(B) = 14$

$P(\text{Sum}=8 | \text{Product}=0) = \frac{1}{14}$ Why this step is important: This is the final calculation that directly answers the problem by combining the results from the previous steps.

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Common Mistakes & Tips

• Sample Space Boundaries: Always remember the defined range of tickets. A common mistake is to consider tickets like $80$ even if they fall outside the $00-49$ range.
• Double Counting: When listing outcomes for $T \times U = 0$, be careful not to double-count tickets like $00$ if you list "tickets with $T=0$" and "tickets with $U=0$" separately without adjustment. A systematic approach (e.g., $T=0$, then $U=0$ for $T \neq 0$) helps avoid this.
• Misinterpreting "Given That": Do not calculate $P(A)$ or $P(B)$ relative to the original sample space $n(S)$ and then try to use $P(A|B) = \frac{P(A \cap B)}{P(B)}$ directly. While correct, it's often simpler to directly work with counts of outcomes ($n(A \cap B)$ and $n(B)$) for the conditional probability when dealing with equally likely events.

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Summary

To find the probability that the sum of the digits is $8$, given that their product is zero, we first identified all tickets where the product of digits is zero (Event B). There were $14$ such tickets. Then, from these $14$ tickets, we identified how many also had a sum of digits equal to $8$ (Event $A \cap B$). Only one ticket, $08$, satisfied both conditions. Therefore, the conditional probability is the ratio of these counts, which is $\frac{1}{14}$.

The final answer is $\boxed{\frac{1}{14}}$, which corresponds to option (A).