A random variable X takes the values 0,1,2,3 and its mean = 1,3 .If P(X=3) =2P(X=1) and P(X=2) = 0.3 then P(X=0) is

Options

A. `0.4`
B. `0.3`
C. `0.2`
D. `0.1`

Answer

To solve the problem step by step, we need to find \( P(X=0) \) given the conditions of the random variable \( X \).

### Step 1: Define the probabilities
Let:
- \( P(X=0) = p_0 \)
- \( P(X=1) = p_1 \)
- \( P(X=2) = p_2 = 0.3 \)
- \( P(X=3) = p_3 \)

From the problem, we know:
- \( P(X=3) = 2P(X=1) \) implies \( p_3 = 2p_1 \)
- The mean of \( X \) is given as \( 1.3 \).

### Step 2: Write the equation for the mean
The mean \( E(X) \) can be calculated using the formula:
\[ E(X) = \sum_{i=0}^{3} i \cdot P(X=i) = 0 \cdot p_0 + 1 \cdot p_1 + 2 \cdot p_2 + 3 \cdot p_3 \]
Substituting the known values:
\[ 1.3 = 0 \cdot p_0 + 1 \cdot p_1 + 2 \cdot 0.3 + 3 \cdot p_3 \]
This simplifies to:
\[ 1.3 = p_1 + 0.6 + 3p_3 \]
Thus,
\[ p_1 + 3p_3 = 1.3 - 0.6 = 0.7 \quad \text{(Equation 1)} \]

### Step 3: Substitute \( p_3 \) in terms of \( p_1 \)
From the relationship \( p_3 = 2p_1 \), substitute this into Equation 1:
\[ p_1 + 3(2p_1) = 0.7 \]
This simplifies to:
\[ p_1 + 6p_1 = 0.7 \implies 7p_1 = 0.7 \]
Thus,
\[ p_1 = \frac{0.7}{7} = 0.1 \]

### Step 4: Find \( p_3 \)
Now substitute \( p_1 \) back to find \( p_3 \):
\[ p_3 = 2p_1 = 2 \cdot 0.1 = 0.2 \]

### Step 5: Use the total probability to find \( p_0 \)
We know that the total probability must equal 1:
\[ p_0 + p_1 + p_2 + p_3 = 1 \]
Substituting the known values:
\[ p_0 + 0.1 + 0.3 + 0.2 = 1 \]
This simplifies to:
\[ p_0 + 0.6 = 1 \implies p_0 = 1 - 0.6 = 0.4 \]

### Conclusion
Thus, the probability \( P(X=0) \) is:
\[ \boxed{0.4} \]