1. **Problem statement:** The survival time after an operation follows an exponential distribution with an average (mean) of 4 years. Given that a patient has already survived 2 years, we want to find the probability that the patient will survive at least 3 more years. 2. **Formula and properties:** For an exponential distribution with parameter $\lambda$, the mean is $\frac{1}{\lambda}$. Here, mean $=4$ years, so: $$\lambda = \frac{1}{4}$$ The exponential distribution has the **memoryless property**, which means: $$P(T > s + t \mid T > s) = P(T > t)$$ 3. **Applying the memoryless property:** Since the patient has already survived 2 years, the probability of surviving at least 3 more years is: $$P(T > 2 + 3 \mid T > 2) = P(T > 3)$$ 4. **Calculating $P(T > 3)$:** The survival function for exponential distribution is: $$P(T > t) = e^{-\lambda t}$$ Substitute $\lambda = \frac{1}{4}$ and $t=3$: $$P(T > 3) = e^{-\frac{1}{4} \times 3} = e^{-\frac{3}{4}}$$ 5. **Final answer:** $$P(\text{survive at least 3 more years} \mid \text{already survived 2 years}) = e^{-\frac{3}{4}} \approx 0.4724$$ This means there is approximately a 47.24% chance the patient will survive at least 3 more years given they have already survived 2 years.