Question

You are rolling two number cubes. Let event A be the sum of the rolled numbers is 7 and event B be the first number is 4. Find the probability P(A and B) and determine whether events A and B are independent.
Enter the first answer as a decimal rounded to the nearest hundredth.
(1 point)
P (A and B) =
Enter 1 if events A and B are
dependent. Enter 2 if events A and B
are independent

Answer 1

To find the probability \( P(A \text{ and } B) \), we need to evaluate both events:

1. **Event A**: The sum of the rolled numbers is 7.
2. **Event B**: The first number rolled is 4.

When rolling two six-sided dice, we can determine the possible outcomes:

The pairs of dice results that sum to 7 are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)

So, there are 6 combinations that satisfy event A.

Now, let's look at event B, where the first die is 4. The possible outcomes for the second die are:
- (4, 1)
- (4, 2)
- (4, 3)
- (4, 4)
- (4, 5)
- (4, 6)

There are 6 outcomes where the first die is 4.

Now we need to find \( P(A \text{ and } B) \), which represents the scenario when both events occur at the same time. We look for outcomes that satisfy both conditions. For event \( B \) to occur (the first die is 4):

The only pair that satisfies both events is (4, 3) since 4 + 3 = 7.

Thus, the outcome that satisfies both \( A \text{ and } B \) is only (4, 3). Therefore, there is exactly **1** outcome in this case.

Now, let’s calculate the total number of possible outcomes when rolling two dice:

The total combinations when rolling two dice = \( 6 \times 6 = 36 \).

Then, we compute:
\[
P(A \text{ and } B) = \frac{\text{Number of favorable outcomes for } A \text{ and } B}{\text{Total possible outcomes}} = \frac{1}{36}.
\]

Now, let’s express \( P(A \text{ and } B) \) as a decimal rounded to the nearest hundredth:
\[
P(A \text{ and } B) \approx 0.0278 \to 0.03 \text{ (rounded to nearest hundredth)}.
\]

Next, to check if events \( A \) and \( B \) are independent, we need to see if:
\[
P(A \text{ and } B) = P(A) \cdot P(B).
\]

1. Calculate \( P(A) \):
\[
P(A) = \frac{6}{36} = \frac{1}{6}.
\]

2. Calculate \( P(B) \):
\[
P(B) = \frac{6}{36} = \frac{1}{6}.
\]

Now, calculate \( P(A) \cdot P(B) \):
\[
P(A) \cdot P(B) = \left( \frac{1}{6} \right) \cdot \left( \frac{1}{6} \right) = \frac{1}{36}.
\]

Since \( P(A \text{ and } B) = \frac{1}{36} \) is equal to \( P(A) \cdot P(B) \), events \( A \) and \( B \) are independent.

Summarizing the results:
\[
P(A \text{ and } B) = 0.03
\]

Enter:
- \( P(A \text{ and } B) = 0.03 \)
- Events A and B are independent, so enter \( 2 \).