1. **Problem statement:** Given a Poisson random variable $X$ such that $P(X=0) = P(X=1)$, find the expected value $E(X)$. 2. **Recall the Poisson distribution formula:** $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ where $\lambda = E(X)$ is the parameter of the distribution. 3. **Apply the given condition:** $$P(X=0) = P(X=1)$$ Substitute the formula: $$\frac{e^{-\lambda} \lambda^0}{0!} = \frac{e^{-\lambda} \lambda^1}{1!}$$ 4. **Simplify the equation:** $$e^{-\lambda} = e^{-\lambda} \lambda$$ Divide both sides by $e^{-\lambda}$ (which is never zero): $$1 = \lambda$$ 5. **Conclusion:** The expected value of $X$ is: $$E(X) = \lambda = 1$$