Key Concepts and Formulas

1. Sum of Probabilities: For any discrete probability distribution, the sum of probabilities of all possible outcomes must equal 1. $\sum P(X=x_i) = 1$

2. Mean (Expected Value): The mean or expected value of a discrete random variable $X$, denoted as $E(X)$ or $\mu$, is calculated by summing the product of each possible value of $X$ and its corresponding probability. $E(X) = \sum x_i P(X=x_i)$

3. Variance: The variance of a discrete random variable $X$, denoted as $Var(X)$ or $\sigma^2$, measures the spread of the distribution. It can be calculated using the formula: $Var(X) = E(X^2) - (E(X))^2$ where $E(X^2) = \sum x_i^2 P(X=x_i)$.

Step-by-Step Solution

Step 1: Use the sum of probabilities to form an equation for 'a' and 'b'. The given probabilities are $a, 2a, a+b, 2b, 3b$. According to the property that the sum of all probabilities must be 1: $a + 2a + (a+b) + 2b + 3b = 1$ Combine like terms: $(a+2a+a) + (b+2b+3b) = 1$ $4a + 6b = 1 \quad \text{(Equation 1)}$

Step 2: Use the given mean to form a second equation for 'a' and 'b'. The mean $E(X)$ is given as $\frac{46}{9}$. We calculate $E(X)$ using the formula $E(X) = \sum x_i P(X=x_i)$: $E(X) = (0 \cdot a) + (2 \cdot 2a) + (4 \cdot (a+b)) + (6 \cdot 2b) + (8 \cdot 3b)$ $E(X) = 0 + 4a + 4a + 4b + 12b + 24b$ Combine like terms: $E(X) = (4a+4a) + (4b+12b+24b)$ $E(X) = 8a + 40b$ We are given $E(X) = \frac{46}{9}$, so: $8a + 40b = \frac{46}{9} \quad \text{(Equation 2)}$

Step 3: Solve the system of linear equations for 'a' and 'b'. We have two equations:

1. $4a + 6b = 1$
2. $8a + 40b = \frac{46}{9}$

Multiply Equation 1 by 2 to eliminate 'a': $2 \cdot (4a + 6b) = 2 \cdot 1$ $8a + 12b = 2 \quad \text{(Equation 3)}$

Subtract Equation 3 from Equation 2: $(8a + 40b) - (8a + 12b) = \frac{46}{9} - 2$ $28b = \frac{46}{9} - \frac{18}{9}$ $28b = \frac{28}{9}$ Divide by 28: $b = \frac{1}{9}$

Substitute the value of $b$ into Equation 1: $4a + 6\left(\frac{1}{9}\right) = 1$ $4a + \frac{2}{3} = 1$ $4a = 1 - \frac{2}{3}$ $4a = \frac{1}{3}$ $a = \frac{1}{12}$ So, we have $a = \frac{1}{12}$ and $b = \frac{1}{9}$.

Step 4: Calculate $E(X^2)$. Now we need to calculate $E(X^2)$ using the formula $E(X^2) = \sum x_i^2 P(X=x_i)$. First, let's list the probabilities with the calculated values of $a$ and $b$:

• $P(X=0) = a = \frac{1}{12}$
• $P(X=2) = 2a = 2 \cdot \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$
• $P(X=4) = a+b = \frac{1}{12} + \frac{1}{9} = \frac{3}{36} + \frac{4}{36} = \frac{7}{36}$
• $P(X=6) = 2b = 2 \cdot \frac{1}{9} = \frac{2}{9}$
• $P(X=8) = 3b = 3 \cdot \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$

Now calculate $E(X^2)$: $E(X^2) = (0^2 \cdot P(X=0)) + (2^2 \cdot P(X=2)) + (4^2 \cdot P(X=4)) + (6^2 \cdot P(X=6)) + (8^2 \cdot P(X=8))$ $E(X^2) = (0 \cdot \frac{1}{12}) + (4 \cdot \frac{1}{6}) + (16 \cdot \frac{7}{36}) + (36 \cdot \frac{2}{9}) + (64 \cdot \frac{1}{3})$ $E(X^2) = 0 + \frac{4}{6} + \frac{112}{36} + \frac{72}{9} + \frac{64}{3}$ Simplify the fractions: $E(X^2) = 0 + \frac{2}{3} + \frac{28}{9} + 8 + \frac{64}{3}$ To sum these, find a common denominator, which is 9: $E(X^2) = \frac{2 \cdot 3}{9} + \frac{28}{9} + \frac{8 \cdot 9}{9} + \frac{64 \cdot 3}{9}$ $E(X^2) = \frac{6}{9} + \frac{28}{9} + \frac{72}{9} + \frac{192}{9}$ $E(X^2) = \frac{6 + 28 + 72 + 192}{9}$ $E(X^2) = \frac{298}{9}$

Step 5: Calculate the Variance. Using the formula $Var(X) = E(X^2) - (E(X))^2$: We have $E(X^2) = \frac{298}{9}$ and $E(X) = \frac{46}{9}$. $(E(X))^2 = \left(\frac{46}{9}\right)^2 = \frac{46^2}{9^2} = \frac{2116}{81}$ Now substitute these values into the variance formula: $Var(X) = \frac{298}{9} - \frac{2116}{81}$ To subtract, find a common denominator, which is 81: $Var(X) = \frac{298 \cdot 9}{81} - \frac{2116}{81}$ $Var(X) = \frac{2682}{81} - \frac{2116}{81}$ $Var(X) = \frac{2682 - 2116}{81}$ $Var(X) = \frac{566}{81}$

Common Mistakes & Tips

• Forgetting $\sum P(X=x_i) = 1$: This is a crucial first step to set up one of the equations for unknown probabilities. Always start here.
• Arithmetic Errors: Calculations involving fractions and squares can be prone to errors. Double-check each step, especially when finding common denominators or squaring fractions.
• Incorrect Variance Formula: A common mistake is to confuse $E(X^2)$ with $(E(X))^2$. Remember that $Var(X) = E(X^2) - (E(X))^2$, not $E((X-E(X))^2)$ which is the definition but less practical for calculation.
• Organizing Calculations: Keep your work neat and organized, especially when dealing with multiple terms in sums for $E(X)$ and $E(X^2)$. Using a table for $x_i$, $P(X=x_i)$, $x_i P(X=x_i)$, $x_i^2$, and $x_i^2 P(X=x_i)$ can help prevent errors.

Summary

We first used the property that the sum of all probabilities in a distribution must equal 1 to form the first linear equation in terms of $a$ and $b$. Next, we calculated the expected value (mean) of the distribution in terms of $a$ and $b$ and equated it to the given mean of $\frac{46}{9}$, yielding a second linear equation. Solving this system of equations gave us the values of $a=\frac{1}{12}$ and $b=\frac{1}{9}$. With these values, we then calculated $E(X^2)$ by summing $x_i^2 P(X=x_i)$ for all values. Finally, we used the variance formula, $Var(X) = E(X^2) - (E(X))^2$, to find the variance of the distribution. Our calculations show the variance to be $\frac{566}{81}$.

The final answer is $\boxed{\frac{566}{81}}$ which corresponds to option (B).