R135 · Conditional dice probability

The total number of outcomes where at least one die shows a 4 (Event $A$) is 11. The outcomes are those pairs $(D_1, D_2)$ where $D_1=4$ or $D_2=4$.

We are interested in the event that the dice sum to 9 (Event $B$). We need to find the outcomes common to both $A$ and $B$, which is $A \cap B$. The pairs summing to 9 are $(3, 6), (4, 5), (5, 4), (6, 3)$. Among these, only $(4, 5)$ and $(5, 4)$ contain a 4.

Thus, $|A \cap B| = 2$. The conditional probability $P(B|A)$ is the ratio of favorable outcomes in the restricted sample space:
$$P(\text{Sum}=9 | A) = \frac{|A \cap B|}{|A|} = \frac{2}{11}$$

**Answer:** $\frac{2}{11}$

The probability is **2/11**.

**Justification:**
The reduced sample space of outcomes where at least one die shows a 4 contains 11 possibilities: (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), and (6,4). Among these 11 outcomes, only two result in a sum of 9: (4,5) and (5,4). Therefore, the probability is the ratio of these successful outcomes to the reduced sample space.

To determine the probability that two fair six-sided dice sum to \(9\) given that at least one die shows a \(4\), follow these steps:

1. **Identify Total Possible Outcomes with At Least One 4:**
   
   When rolling two dice, there are \(6 \times 6 = 36\) possible outcomes.
   
   - The number of outcomes where the first die is \(4\) (regardless of the second die) is \(6\).
   - Similarly, the number of outcomes where the second die is \(4\) is also \(6\).
   - However, the outcome where both dice are \(4\) is counted twice in the above counts. To correct for this:
     
     \[
     \text{Total outcomes with at least one } 4 = 6 + 6 - 1 = 11
     \]

2. **Determine Favorable Outcomes (Sum to 9):**
   
   We need pairs \((a, b)\) such that \(a + b = 9\) and at least one of \(a\) or \(b\) is \(4\).
   
   - The possible pairs are:
     \[
     (4, 5) \quad \text{and} \quad (5, 4)
     \]
     
   So, there are **2** favorable outcomes.

3. **Calculate the Probability:**
   
   \[
   P(\text{Sum }9 | \text{At least one die is }4) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes with At Least One 4}} = \frac{2}{11}
   \]

**Final Answer:**

\[
\boxed{\dfrac{2}{11}}
\]

The probability that the two dice sum to 9, given that at least one die shows a 4, is $\frac{2}{11}$.

**Justification:**
Let $A$ be the event that at least one die shows a 4. The outcomes for $A$ are $(4, x)$ or $(x, 4)$, excluding duplicates; there are 11 such outcomes (e.g., (4, 1), (1, 4), ..., (4, 6), (6, 4)). Let $B$ be the event that the sum is 9. The outcomes for $A \cap B$ must satisfy both conditions: they must sum to 9 and contain at least one 4. These outcomes are $(4, 5)$ and $(5, 4)$. Since there are 2 favorable outcomes out of a total of 11 possible outcomes in the reduced sample space $A$, the conditional probability is $\frac{2}{11}$.

Let $A$ be the event that at least one die shows a 4, and $B$ be the event that the two dice sum to 9. We need to find $P(B | A) = \frac{|A \cap B|}{|A|}$.

The total number of outcomes is 36. The size of event $A$ (at least one die is 4) is $|A| = 11$.
The outcomes where the sum is 9 are $\{(3, 6), (4, 5), (5, 4), (6, 3)\}$.
We find the intersection $A \cap B$: these outcomes must sum to 9 and include at least one 4. The outcomes satisfying both conditions are $(4, 5)$ and $(5, 4)$. Thus, $|A \cap B| = 2$.

The probability is $\frac{2}{11}$.

## Step 1: Calculate the total number of outcomes when rolling two dice.
When rolling two six-sided dice, there are $6 \times 6 = 36$ possible outcomes.

## Step 2: Determine the number of outcomes where at least one die shows a 4.
We can break this down into cases. The first case is if the first die shows a 4 and the second does not (with 5 possibilities for the second die). The second case is if the first die does not show a 4 but the second does (also with 5 possibilities). However, we can simplify by considering that regardless of which die shows the 4, there are a total of $2 \times 5 = 10$ outcomes where at least one die shows a 4.

## Step 3: Determine the number of outcomes where both dice sum to 9.
To get a sum of 9 with two six-sided dice, we can have (3,6), (4,5), (5,4), or (6,3). This gives us exactly 4 successful outcomes.

## Step 4: Calculate the probability using the given condition that at least one die shows a 4.
The probability of rolling a sum of 9, given that at least one die shows a 4, is the number of successful outcomes divided by the total number of possible outcomes under the given condition. This results in $\frac{2}{10}$.

## Step 5: Simplify the fraction to find the exact answer.
$\frac{2}{10} = \frac{1}{5}$

The final answer is: $\boxed{\frac{1}{5}}$

The probability is **2/11**. There are 11 outcomes where at least one die shows a 4. Among these, only two outcomes—(4,5) and (5,4)—result in a sum of 9. Thus, the conditional probability is 2/11. 

**Answer:** 2/11  
**Justification:** Out of 11 possible outcomes with at least one 4, only two sum to 9.

To solve this problem, we need to determine the probability that the sum of two dice is 9, given that at least one die shows a 4.

First, let's identify all possible outcomes where at least one die shows a 4:
- (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
- (1,4), (2,4), (3,4), (5,4), (6,4)

There are 11 such outcomes.

Next, we need to find which of these outcomes result in a sum of 9:
- (4,5) and (5,4)

There are 2 favorable outcomes.

Therefore, the probability that the two dice sum to 9, given that at least one die shows a 4, is:

\[ \frac{2}{11} \]

The probability of this scenario occurring is 1/3, or 0.33 recurring. 

There are three possible cases: one die shows a 4 while the other has a value of either 5 or 6; both dice show a value of 4; and finally, again with one die showing a 4 but the other having a value of 3. This gives us two favorable outcomes out of six equally likely possibilities in total. Thus, the probability is 2/6, which reduces to 1/3.

The probability is \(\boxed{\dfrac{2}{11}}\).

**Justification:**  
Given at least one die shows a 4 (11 possible outcomes: (4,1) to (4,6) and (1,4) to (6,4), excluding duplicate (4,4)), only two outcomes sum to 9: (4,5) and (5,4). Thus, the probability is \( \frac{2}{11} \).

To determine the probability that two dice sum to **9** given that at least one die shows a **4**, follow these steps:

1. **Total Possible Outcomes:**
   - When rolling two six-sided dice, there are \(6 \times 6 = 36\) possible outcomes.

2. **Favorable Outcomes (Sum to 9):**
   - The pairs that sum to **9** are:
     \[
     (3, 6),\ (4, 5),\ (5, 4),\ (6, 3)
     \]
   - So, there are **4** favorable outcomes.

3. **Outcomes with At Least One Die Showing a 4:**
   - The pairs where at least one die is **4** are:
     \[
     (1, 4),\ (2, 4),\ (3, 4),\ (4, 1),\ (4, 2),\ (4, 3),\ (4, 4)
     \]
   - This gives us **7** possible outcomes.

4. **Intersection of Both Conditions:**
   - From the favorable outcomes that sum to **9**, only two pairs meet the condition of having at least one die showing a **4**:
     \[
     (4, 5),\ (5, 4)
     \]
   - So, there are **2** such outcomes.

5. **Calculating the Probability:**
   \[
   \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes with At Least One 4}} = \frac{2}{7}
   \]

\(\boxed{\dfrac{2}{7}}\)