1. The problem states that $X$ is an exponential random variable with rate parameter $\lambda = \frac{4}{3}$. We want to find the probability $P(X > 1)$, which is the probability that $X$ exceeds 1. 2. The probability density function (pdf) of an exponential distribution is given by: $$f(x) = \lambda e^{-\lambda x}$$ For $\lambda = \frac{4}{3}$, this becomes: $$f(x) = \frac{4}{3} e^{-\frac{4}{3} x}$$ 3. The probability $P(X > 1)$ is the integral of the pdf from 1 to infinity: $$P(X > 1) = \int_1^{\infty} \frac{4}{3} e^{-\frac{4}{3} x} \, dx$$ 4. To solve the integral, we use the antiderivative of the exponential function: $$\int e^{-ax} dx = -\frac{1}{a} e^{-ax} + C$$ Applying this, we get: $$\int_1^{\infty} \frac{4}{3} e^{-\frac{4}{3} x} dx = \left[-e^{-\frac{4}{3} x}\right]_1^{\infty}$$ 5. Evaluating the limits: - As $x \to \infty$, $e^{-\frac{4}{3} x} \to 0$ - At $x=1$, $e^{-\frac{4}{3} \cdot 1} = e^{-\frac{4}{3}}$ So, $$P(X > 1) = 0 - (-e^{-\frac{4}{3}}) = e^{-\frac{4}{3}}$$ 6. Numerically, $e^{-\frac{4}{3}} \approx 0.2636$. 7. Alternatively, since the exponential distribution is memoryless, $P(X > 1) = 1 - P(X \leq 1)$, and $P(X \leq 1) = 1 - e^{-\frac{4}{3}}$, confirming the result. Final answer: $$P(X > 1) = e^{-\frac{4}{3}} \approx 0.2636$$