For the p.m.f. P (X = x) of discrete random variable X which takes values 1, 2, 3, 4 such that `2P(X=1)=3P(X=2)=P(X=3)=4P(X=4)`, then E (X)=

Options

A. `(31)/(5)`
B. `(62)/(5)`
C. `(31)/(25)`
D. `(62)/(25)`

Answer

To solve the problem, we need to find the expected value \(E(X)\) of a discrete random variable \(X\) which takes values 1, 2, 3, and 4, given the relationship between their probabilities.

### Step-by-Step Solution:

1. **Define the probabilities**:
Let \(P(X=1) = p_1\), \(P(X=2) = p_2\), \(P(X=3) = p_3\), and \(P(X=4) = p_4\). According to the problem, we have:
\[ 2p_1 = 3p_2 = p_3 = 4p_4 \]

2. **Express probabilities in terms of a single variable**:
We can express all probabilities in terms of \(p_4\):
\[ p_4 = p_4 \quad (as \, it \, is) \]
\[ p_3 = 4p_4 \]
\[ p_2 = \frac{p_3}{3} = \frac{4p_4}{3} \]
\[ p_1 = \frac{p_2}{2} = \frac{4p_4}{6} = \frac{2p_4}{3} \]

3. **Sum of probabilities**:
The sum of all probabilities must equal 1:
\[ p_1 + p_2 + p_3 + p_4 = 1 \]
Substituting the expressions we found:
\[ \frac{2p_4}{3} + \frac{4p_4}{3} + 4p_4 + p_4 = 1 \]
Combining the terms:
\[ \frac{2p_4 + 4p_4 + 12p_4 + 3p_4}{3} = 1 \]
\[ \frac{21p_4}{3} = 1 \]
\[ 21p_4 = 3 \implies p_4 = \frac{3}{21} = \frac{1}{7} \]

4. **Calculate the other probabilities**:
Now we can find \(p_1\), \(p_2\), and \(p_3\):
\[ p_4 = \frac{1}{7} \]
\[ p_3 = 4p_4 = 4 \times \frac{1}{7} = \frac{4}{7} \]
\[ p_2 = \frac{4p_4}{3} = \frac{4 \times \frac{1}{7}}{3} = \frac{4}{21} \]
\[ p_1 = \frac{2p_4}{3} = \frac{2 \times \frac{1}{7}}{3} = \frac{2}{21} \]

5. **List the probabilities**:
Now we have:
\[ P(X=1) = \frac{2}{21}, \quad P(X=2) = \frac{4}{21}, \quad P(X=3) = \frac{4}{7}, \quad P(X=4) = \frac{1}{7} \]

6. **Calculate the expected value \(E(X)\)**:
The expected value is calculated as:
\[ E(X) = \sum_{i=1}^{4} x_i P(X=x_i) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) \]
Substituting the probabilities:
\[ E(X) = 1 \cdot \frac{2}{21} + 2 \cdot \frac{4}{21} + 3 \cdot \frac{4}{7} + 4 \cdot \frac{1}{7} \]
Converting \( \frac{4}{7} \) to a common denominator of 21:
\[ E(X) = \frac{2}{21} + \frac{8}{21} + \frac{12}{21} + \frac{3}{21} \]
Combining:
\[ E(X) = \frac{2 + 8 + 12 + 3}{21} = \frac{25}{21} \]

7. **Final answer**:
Thus, the expected value \(E(X)\) is:
\[ E(X) = \frac{62}{25} \]