Question

You are rolling two number cubes. Let event A
𝐴
be the sum of the rolled numbers is 7 and event B
𝐵
be the first number is 4.

Find the probability P(A and B)
𝑃
(
𝐴

𝑎
𝑛
𝑑

𝐵
)
and determine whether events A
𝐴
and B
𝐵
are independent.

Enter the first answer as a decimal rounded to the nearest hundredth.

(2 points)
​P(A and B)​
​𝑃
(
𝐴

𝑎
𝑛
𝑑

𝐵
)
​
=

Enter 1 if events A
𝐴
and B
𝐵
are dependent. Enter 2 if events A
𝐴
and B
𝐵
are independent. :

Answer 1

To find \( P(A \text{ and } B) \) where:

- Event \( A \): The sum of the rolled numbers is 7.
- Event \( B \): The first number is 4.

First, let's identify the outcomes for each event.

### Step 1: Identify Event B
Event \( B \) occurs when the first die shows 4. The possible outcomes when rolling two number cubes are:
1. (4, 1)
2. (4, 2)
3. (4, 3)
4. (4, 4)
5. (4, 5)
6. (4, 6)

So, there are 6 possible outcomes where event \( B \) is true.

### Step 2: Identify Event A and A and B
Event \( A \) occurs when the sum of the two dice is 7. The possible pairs of outcomes that give a sum of 7 are:
1. (1, 6)
2. (2, 5)
3. (3, 4)
4. (4, 3)
5. (5, 2)
6. (6, 1)

Now, \( A \) and \( B \) occur together (i.e., \( P(A \text{ and } B) \)) when the first die is 4, and the sum of the two dice equals 7. The only pair matching these criteria is:
- (4, 3)

Thus, \( P(A \text{ and } B) \) corresponds to the event that occurs when the outcomes is (4, 3).

### Step 3: Calculate \( P(A \text{ and } B) \)
There is 1 outcome for \( A \text{ and } B \) which is (4, 3). The total number of outcomes when rolling two dice is \( 6 \times 6 = 36 \). Therefore:

\[
P(A \text{ and } B) = \frac{\text{Number of outcomes of } (A \text{ and } B)}{\text{Total outcomes}} = \frac{1}{36} \approx 0.0278
\]

Rounding to the nearest hundredth, we have:

\[
P(A \text{ and } B) \approx 0.03
\]

### Step 4: Determine Independence of A and B
Events \( A \) and \( B \) are independent if:

\[
P(A \text{ and } B) = P(A) \times P(B)
\]

#### Calculate \( P(A) \)
There are 6 outcomes where the sum is 7:
\[
P(A) = \frac{6}{36} = \frac{1}{6}
\]

#### Calculate \( P(B) \)
There are 6 outcomes where the first die is 4:
\[
P(B) = \frac{6}{36} = \frac{1}{6}
\]

#### Calculate \( P(A) \times P(B) \)
\[
P(A) \times P(B) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36}
\]

### Step 5: Compare
Since \( P(A \text{ and } B) = \frac{1}{36} \) and \( P(A) \times P(B) = \frac{1}{36} \), we find that:

\[
P(A \text{ and } B) = P(A) \times P(B)
\]

This indicates that events \( A \) and \( B \) are independent.

### Final Answers
- \( P(A \text{ and } B) = 0.03 \)
- Events \( A \) and \( B \) are independent, so the answer is \( 2 \).

So the final submission would be:
1. \( P(A \text{ and } B) = 0.03 \)
2. Events A and B are independent (enter 2).