Chapter 3 HW- Probability Topics Current score: AI. Category: Homework 68.95/100 points d. Which of the following explains what " $$ P(A \mid B) $$ " represents? A $$ P(A \mid B) $$ is the probability that the sum of rolling two dice is at most ten given the fact that we have a one or three on the first die, followed by an odd number on the second die. B $$ P(A \mid B) $$ is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at least ten. C $$ P(A \mid B) $$ is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at most ten. D None of the above Find $$ P(A \mid B) $$. $$ P(A \mid B)= $$ Enter your answer e. Are $$ A $$ and $$ B $$ mutually exclusive events? It is possible to have a one or a three on the first die rolling followed by an odd number on the second die when the sum of these two numbers is A at most ten. Numerically, we have $$ P(A $$ AND $$ B)=\frac{1}{6} $$, so the events are not mutually exclusive. It is impossible to have a one or a three on the first die rolling followed by an odd number on the second die when the sum of these two numbers $$ B $$ is at most ten. Numerically, we have $$ P(A $$ AND $$ B)=0 $$, so the events are mutually exclusive.

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UpStudy AI · Accepted Answer

Sure, let's work through **Problem d** and **Problem e** step by step.

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## **Problem d**

**Question:**
Which of the following explains what $$ P(A \mid B) $$ represents?

**Options:**
- **A.** $$ P(A \mid B) $$ is the probability that the sum of rolling two dice is at most ten given the fact that we have a one or three on the first die, followed by an odd number on the second die.
- **B.** $$ P(A \mid B) $$ is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at least ten.
- **C.** $$ P(A \mid B) $$ is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at most ten.
- **D.** None of the above.

**Understanding $$ P(A \mid B) $$:**
The notation $$ P(A \mid B) $$ denotes the **probability of event $$ A $$** occurring **given that event $$ B $$** has occurred.

Let's define the events based on the options:

- **Event $$ A $$:** The sum of rolling two dice is at most ten.
- **Event $$ B $$:** Rolling a one or three on the first die, followed by an odd number on the second die.

**Interpreting the Options:**

- **Option A** correctly states that $$ P(A \mid B) $$ is the probability that the sum is at most ten **given** that $$ B $$ has occurred.
- **Option B** incorrectly reverses the conditions, making it $$ P(B \mid A) $$.
- **Option C** misstates $$ A $$ and $$ B $$ within the conditional probability.
- **Option D** is incorrect because **Option A** is accurate.

**Conclusion:**
**Option A** is the correct interpretation of $$ P(A \mid B) $$.

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**Calculating $$ P(A \mid B) $$:**

The formula for conditional probability is:

$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
$$

**Step 1: Calculate $$ P(B) $$**

- Rolling a one or three on the first die: $$ 2/6 = 1/3 $$
- Rolling an odd number on the second die: $$ 3/6 = 1/2 $$
- Since the two dice are independent:

$$
P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}
$$

**Step 2: Calculate $$ P(A \cap B) $$**

- $$ A \cap B $$ means both:
  - The sum is at most ten.
  - The first die is a one or three, and the second die is odd.

First, list all possible outcomes for $$ B $$:
- First die: 1 or 3
- Second die: 1, 3, or 5
- Total favorable outcomes for $$ B $$: $$ 2 \times 3 = 6 $$

Now, check which of these have a sum at most ten:
- $$ (1,1) $$: $$ 1 + 1 = 2 $$ ✓
- $$ (1,3) $$: $$ 1 + 3 = 4 $$ ✓
- $$ (1,5) $$: $$ 1 + 5 = 6 $$ ✓
- $$ (3,1) $$: $$ 3 + 1 = 4 $$ ✓
- $$ (3,3) $$: $$ 3 + 3 = 6 $$ ✓
- $$ (3,5) $$: $$ 3 + 5 = 8 $$ ✓

All outcomes in $$ B $$ also satisfy $$ A $$ (since the maximum sum here is 8, which is less than 10).

$$
P(A \cap B) = P(B) = \frac{1}{6}
$$

**Step 3: Compute $$ P(A \mid B) $$**

$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{6}}{\frac{1}{6}} = 1
$$

**Answer:**
$$
P(A \mid B) = 1
$$

This means that if you have a one or three on the first die followed by an odd number on the second die, the sum of the two dice will **always** be at most ten.

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## **Problem e**

**Question:**
Are $$ A $$ and $$ B $$ mutually exclusive events?

**Options Explanation:**

1. **First Statement:**
   - It is possible to have a one or a three on the first die followed by an odd number on the second die **when the sum of these two numbers is at most ten**.
   - Numerically, $$ P(A \text{ AND } B) = \frac{1}{6} $$, so the events are **not** mutually exclusive.

2. **Second Statement:**
   - It is **impossible** to have a one or a three on the first die followed by an odd number on the second die **when the sum of these two numbers is at most ten**.
   - Numerically, $$ P(A \text{ AND } B) = 0 $$, so the events are **mutually exclusive**.

**Determining Mutual Exclusivity:**

Two events $$ A $$ and $$ B $$ are **mutually exclusive** if they **cannot occur simultaneously**, i.e., $$ P(A \cap B) = 0 $$.

From **Problem d**, we found:
$$
P(A \cap B) = \frac{1}{6} \neq 0
$$

**Conclusion:**
Since $$ P(A \cap B) = \frac{1}{6} $$, which is not zero, $$ A $$ and $$ B $$ are **not mutually exclusive**.

**Answer:**
- **$$ A $$ and $$ B $$ are not mutually exclusive events** because there is a non-zero probability ($$ \frac{1}{6} $$) that both events occur together.

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## **Summary**

- **Problem d:** 
  - **Correct Option:** A
  - **$$ P(A \mid B) = 1 $$**
  
- **Problem e:**
  - **$$ A $$ and $$ B $$ are not mutually exclusive events.**

Feel free to reach out if you need further clarification or assistance!
**Problem d:** - **Answer:** Option A is correct. $$ P(A \mid B) $$ is the probability that the sum of rolling two dice is at most ten given that the first die is a one or three and the second die is odd. - **Calculation:** $$ P(A \mid B) = 1 $$ **Problem e:** - **Answer:** $$ A $$ and $$ B $$ are not mutually exclusive events because there is a non-zero probability ($$ \frac{1}{6} $$) that both events occur together.

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