Question:
A random variable X takes values 0, 1, 2, 3 and its mean is 1.3. If $P(X=3)=2P(X=1)$ and $P(X=2)=0.3$, then find $P(X=0)$.
A random variable X takes values 0, 1, 2, 3 and its mean is 1.3. If $P(X=3)=2P(X=1)$ and $P(X=2)=0.3$, then find $P(X=0)$.
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Always start by checking if the sum of probabilities is 1; this often provides the first necessary constraint to solve for unknown variables.
Updated On: Jun 9, 2026
- \(\frac{1}{5} \)
- \(\frac{2}{5} \)
- \(\frac{3}{5} \)
- \(\frac{4}{5} \)
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The Correct Option is B
Solution and Explanation
Concept:
The expected value (mean) is defined as $E[X] = \sum x_i P(x_i)$, and the sum of all probabilities must equal 1.
Step 1: Express the probabilities using variables.
Let $P(X=1) = p$. Given $P(X=3) = 2p$ and $P(X=2) = 0.3$.
Let $P(X=0) = p_0$.
Step 2: Formulate the equation from the sum of probabilities.
\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 \] \[ p_0 + p + 0.3 + 2p = 1 \implies p_0 + 3p = 0.7 \quad \text{--- (Eq. 1)} \]
Step 3: Formulate the equation from the definition of the mean.
\[ E[X] = (0 \cdot p_0) + (1 \cdot p) + (2 \cdot 0.3) + (3 \cdot 2p) = 1.3 \] \[ p + 0.6 + 6p = 1.3 \implies 7p = 0.7 \implies p = 0.1 \]
Step 4: Solve for \(p_0\).
Substitute $p=0.1$ into Eq. 1:
\[ p_0 + 3(0.1) = 0.7 \implies p_0 + 0.3 = 0.7 \implies p_0 = 0.4 \] \[ p_0 = \frac{4}{10} = \frac{2}{5} \] center minipage0.3
P(X=0) = 2/5 minipage center
Step 1: Express the probabilities using variables.
Let $P(X=1) = p$. Given $P(X=3) = 2p$ and $P(X=2) = 0.3$.
Let $P(X=0) = p_0$.
Step 2: Formulate the equation from the sum of probabilities.
\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 \] \[ p_0 + p + 0.3 + 2p = 1 \implies p_0 + 3p = 0.7 \quad \text{--- (Eq. 1)} \]
Step 3: Formulate the equation from the definition of the mean.
\[ E[X] = (0 \cdot p_0) + (1 \cdot p) + (2 \cdot 0.3) + (3 \cdot 2p) = 1.3 \] \[ p + 0.6 + 6p = 1.3 \implies 7p = 0.7 \implies p = 0.1 \]
Step 4: Solve for \(p_0\).
Substitute $p=0.1$ into Eq. 1:
\[ p_0 + 3(0.1) = 0.7 \implies p_0 + 0.3 = 0.7 \implies p_0 = 0.4 \] \[ p_0 = \frac{4}{10} = \frac{2}{5} \] center minipage0.3
P(X=0) = 2/5 minipage center
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