If X is a poisson variate such that `P(X=0)=P(X=1)=K`, then show that K = 1/e

Answer

To solve the problem, we start with the given information that \( P(X=0) = P(X=1) = K \) for a Poisson random variable \( X \). We need to show that \( K = \frac{1}{e} \).

### Step-by-Step Solution:

1. **Recall the Poisson Probability Formula**:
The probability mass function for a Poisson random variable \( X \) with parameter \( \lambda \) is given by:
\[ P(X = n) = \frac{\lambda^n e^{-\lambda}}{n!} \]

2. **Set Up the Equations**:
According to the problem, we have:
\[ P(X = 0) = K \quad \text{and} \quad P(X = 1) = K \]
Using the Poisson formula, we can express these probabilities:
- For \( n = 0 \):
\[ P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} \]
- For \( n = 1 \):
\[ P(X = 1) = \frac{\lambda^1 e^{-\lambda}}{1!} = \lambda e^{-\lambda} \]

3. **Equate the Two Probabilities**:
Since both probabilities equal \( K \), we can set them equal to each other:
\[ e^{-\lambda} = \lambda e^{-\lambda} \]

4. **Simplify the Equation**:
We can divide both sides by \( e^{-\lambda} \) (note that \( e^{-\lambda} \neq 0 \)):
\[ 1 = \lambda \]

5. **Find the Value of \( \lambda \)**:
From the above equation, we find:
\[ \lambda = 1 \]

6. **Substitute \( \lambda \) Back to Find \( K \)**:
Now we can substitute \( \lambda = 1 \) back into the expression for \( K \):
\[ K = P(X = 0) = e^{-\lambda} = e^{-1} = \frac{1}{e} \]

### Conclusion:
Thus, we have shown that:
\[ K = \frac{1}{e} \]