To solve the problem step-by-step, we will find the probability distribution of the number of successes when a die is tossed twice, where 'getting a number greater than 4' is considered a success. We will also calculate the mean and variance of the number of successes.
### Step 1: Determine the Sample Space
When a die is tossed twice, the total number of outcomes is:
\[
6 \times 6 = 36
\]
The sample space consists of all possible pairs of outcomes from the two tosses, which can be represented as (1,1), (1,2), ..., (6,6).
### Step 2: Define Success
A success is defined as getting a number greater than 4. The numbers greater than 4 on a die are 5 and 6. Therefore, the successful outcomes for each die toss are:
- Outcomes greater than 4: 5, 6
### Step 3: Calculate the Probability of Successes
We will calculate the probability of getting 0, 1, or 2 successes (i.e., the number of times we get a number greater than 4 in two tosses).
- **0 successes**: This occurs when both tosses result in numbers less than or equal to 4. The outcomes for this are (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4).
- Total outcomes: 16
- Probability \( P(0) = \frac{16}{36} = \frac{4}{9} \)
- **1 success**: This occurs when one toss results in a number greater than 4 and the other does not. The successful pairs are:
- (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (1,5), (2,5), (3,5), (4,5), (1,6), (2,6), (3,6), (4,6).
- Total outcomes: 12
- Probability \( P(1) = \frac{12}{36} = \frac{1}{3} \)
- **2 successes**: This occurs when both tosses result in numbers greater than 4. The successful pairs are:
- (5,5), (5,6), (6,5), (6,6).
- Total outcomes: 4
- Probability \( P(2) = \frac{4}{36} = \frac{1}{9} \)
### Step 4: Probability Distribution Table
Now we can summarize the probabilities in a table:
| Number of Successes (x) | Probability P(x) |
|-------------------------|-------------------|
| 0 | \( \frac{4}{9} \) |
| 1 | \( \frac{1}{3} \) |
| 2 | \( \frac{1}{9} \) |
### Step 5: Calculate the Mean
The mean (expected value) \( E(X) \) is calculated using the formula:
\[
E(X) = \sum (x \cdot P(x))
\]
Calculating:
\[
E(X) = 0 \cdot \frac{4}{9} + 1 \cdot \frac{1}{3} + 2 \cdot \frac{1}{9}
\]
\[
= 0 + \frac{1}{3} + \frac{2}{9}
\]
To add \( \frac{1}{3} \) and \( \frac{2}{9} \), convert \( \frac{1}{3} \) to a fraction with a denominator of 9:
\[
\frac{1}{3} = \frac{3}{9}
\]
Thus,
\[
E(X) = \frac{3}{9} + \frac{2}{9} = \frac{5}{9}
\]
### Step 6: Calculate the Variance
The variance \( Var(X) \) is calculated using the formula:
\[
Var(X) = E(X^2) - (E(X))^2
\]
First, we need to calculate \( E(X^2) \):
\[
E(X^2) = \sum (x^2 \cdot P(x))
\]
Calculating:
\[
E(X^2) = 0^2 \cdot \frac{4}{9} + 1^2 \cdot \frac{1}{3} + 2^2 \cdot \frac{1}{9}
\]
\[
= 0 + \frac{1}{3} + \frac{4}{9}
\]
Convert \( \frac{1}{3} \) to a fraction with a denominator of 9:
\[
\frac{1}{3} = \frac{3}{9}
\]
Thus,
\[
E(X^2) = \frac{3}{9} + \frac{4}{9} = \frac{7}{9}
\]
Now, substituting back into the variance formula:
\[
Var(X) = E(X^2) - (E(X))^2 = \frac{7}{9} - \left(\frac{5}{9}\right)^2
\]
Calculating \( \left(\frac{5}{9}\right)^2 = \frac{25}{81} \):
\[
Var(X) = \frac{7}{9} - \frac{25}{81}
\]
Convert \( \frac{7}{9} \) to a fraction with a denominator of 81:
\[
\frac{7}{9} = \frac{63}{81}
\]
Thus,
\[
Var(X) = \frac{63}{81} - \frac{25}{81} = \frac{38}{81}
\]
### Final Results
- **Probability Distribution**:
- \( P(0) = \frac{4}{9} \)
- \( P(1) = \frac{1}{3} \)
- \( P(2) = \frac{1}{9} \)
- **Mean**: \( E(X) = \frac{5}{9} \)
- **Variance**: \( Var(X) = \frac{38}{81} \)
