To find \( P(A \text{ and } B) \), we first need to understand what events A and B mean in the context of rolling two dice.
- Event A: The sum of the rolled numbers is 7.
- Event B: The first number is 4.
First, we can list the pairs of rolls that would satisfy event A (sum = 7):
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)
So, there are 6 outcomes that satisfy event A.
Next, we need to consider when both events A and B happen simultaneously (A and B). For event B to occur (where the first die shows a 4), we only need to look for the outcomes where the first die is 4 and the total sum is 7. The only roll that satisfies this is:
- (4, 3) (the second die must be 3 to make the total 7)
So, there is only 1 outcome that satisfies both events A and B.
Now, we calculate the probability:
There are a total of \(6 \times 6 = 36\) possible outcomes when rolling two number cubes.
Thus, the probability \( P(A \text{ and } B) \) is:
\[ P(A \text{ and } B) = \frac{\text{Number of outcomes that satisfy both A and B}}{\text{Total number of outcomes}} = \frac{1}{36} \approx 0.02778 \]
Rounding to the nearest hundredth gives:
\[ P(A \text{ and } B) \approx 0.03 \]
Next, we check whether events A and B are independent. To determine independence, we need to check if:
\[ P(A \text{ and } B) = P(A) \cdot P(B) \]
First, let’s calculate \( P(A) \) and \( P(B) \):
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\( P(A) \): The total outcomes for event A (sum = 7) is 6. So, \[ P(A) = \frac{6}{36} = \frac{1}{6} \]
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\( P(B) \): The first die shows 4. There are 6 possible outcomes for the second die, so: \[ P(B) = \frac{6}{36} = \frac{1}{6} \]
Now we compute \( P(A) \cdot P(B) \): \[ P(A) \cdot P(B) = \left(\frac{1}{6}\right) \cdot \left(\frac{1}{6}\right) = \frac{1}{36} \]
Since \( P(A \text{ and } B) = \frac{1}{36} \) and \( P(A) \cdot P(B) = \frac{1}{36} \), we conclude that the events A and B are independent.
Therefore, the final answers are:
- \( P(A \text{ and } B) = 0.03 \)
- Events A and B are independent: enter 2.
38 answers