To find $P(X > 1, Y > 4)$ for two independent exponential random variables $X$ and $Y$ with rate $\lambda = 2$, we proceed as follows:

1. Probability for Each Individual Variable:

   The probability that an exponential random variable $X$ with rate $\lambda$ exceeds a certain value $a$ is: $P(X > a) = e^{-\lambda a}$

2. Calculate Each Probability Individually:

   For $X > 1$: $P(X > 1) = e^{-2 \cdot 1} = e^{-2}$

   For $Y > 4$: $P(Y > 4) = e^{-2 \cdot 4} = e^{-8}$

3. Joint Probability (Independence):

   Since $X$ and $Y$ are independent, we can multiply the probabilities: $P(X > 1, Y > 4) = P(X > 1) \cdot P(Y > 4) = e^{-2} \cdot e^{-8} = e^{-10}$

4. Calculate $e^{-10}$:

   $e^{-10} \approx 0.0000454$

Final Answer:

$P(X > 1, Y > 4) \approx 0.000$

Would you like further details on the calculation steps or any additional questions?

Here are some related questions:

1. How would the probability change if $\lambda$ were different?
2. What if $X$ and $Y$ were not independent?
3. How can we generalize this approach to other distributions?
4. What if we want $P(X > 1 \text{ or } Y > 4)$?
5. How does the mean of an exponential distribution affect probabilities over a threshold?

Tip: For exponential distributions, the probability $P(X > x) = e^{-\lambda x}$ simplifies calculations involving thresholds!