If X follows Poisson distribution such that P (X = 1) = 0.4 and P (X = 2) = 0.2, Then find the value of `m`

Answer

To solve the problem, we need to find the value of \( m \) for a Poisson distribution given the probabilities \( P(X = 1) = 0.4 \) and \( P(X = 2) = 0.2 \).

### Step-by-step Solution:

1. **Understand the Poisson Probability Formula**:
The probability mass function for a Poisson distribution is given by:
\[ P(X = k) = \frac{e^{-m} m^k}{k!} \]
where \( m \) is the mean (or parameter) of the distribution.

2. **Set Up the Equations**:
From the problem, we have:
- For \( P(X = 1) = 0.4 \):
\[ P(X = 1) = \frac{e^{-m} m^1}{1!} = 0.4 \]
This simplifies to:
\[ e^{-m} m = 0.4 \quad \text{(Equation 1)} \]

- For \( P(X = 2) = 0.2 \):
\[ P(X = 2) = \frac{e^{-m} m^2}{2!} = 0.2 \]
This simplifies to:
\[ e^{-m} \frac{m^2}{2} = 0.2 \quad \text{(Equation 2)} \]

3. **Substitute Equation 1 into Equation 2**:
From Equation 1, we know:
\[ e^{-m} = \frac{0.4}{m} \]
Substitute this into Equation 2:
\[ \frac{0.4}{m} \cdot \frac{m^2}{2} = 0.2 \]
Simplifying this gives:
\[ \frac{0.4 m}{2} = 0.2 \]
Thus:
\[ 0.2 m = 0.2 \]

4. **Solve for \( m \)**:
Dividing both sides by \( 0.2 \):
\[ m = 1 \]

5. **Conclusion**:
The value of \( m \) is \( 1 \).