1. **Problem statement:** Given a Poisson random variable $X$ such that $P[X=2] = \frac{2}{3} P[X=1]$, find $P[X=3]$. 2. **Recall the Poisson probability mass function (pmf):** $$P[X=k] = \frac{e^{-\lambda} \lambda^k}{k!}$$ where $\lambda$ is the mean (and variance) of the Poisson distribution. 3. **Express the given condition using the pmf:** $$P[X=2] = \frac{2}{3} P[X=1]$$ Substitute the pmf: $$\frac{e^{-\lambda} \lambda^2}{2!} = \frac{2}{3} \times \frac{e^{-\lambda} \lambda^1}{1!}$$ 4. **Simplify the equation:** Cancel $e^{-\lambda}$ on both sides: $$\frac{\lambda^2}{2} = \frac{2}{3} \lambda$$ Multiply both sides by 6 to clear denominators: $$3 \lambda^2 = 4 \lambda$$ Divide both sides by $\lambda$ (assuming $\lambda > 0$): $$3 \lambda = 4$$ 5. **Solve for $\lambda$:** $$\lambda = \frac{4}{3}$$ 6. **Find $P[X=3]$ using the pmf:** $$P[X=3] = \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\frac{4}{3}} \left(\frac{4}{3}\right)^3}{6}$$ 7. **Final answer:** $$P[X=3] = \frac{e^{-\frac{4}{3}} \left(\frac{64}{27}\right)}{6} = \frac{64}{162} e^{-\frac{4}{3}} = \frac{32}{81} e^{-\frac{4}{3}}$$ This is the probability $P[X=3]$ for the given Poisson distribution.