The lifetime X of a light bulb is a random variable with P[X > t] = 2/(2+t) for
t>0. Suppose two light bulbs are installed at time t=0. Given at time t = 1, both
light bulbs are still working, find the conditional probability that at least one light
bulb is still working at time t = 9

To find the conditional probability that at least one light bulb is still working at time \( t = 9 \) given that both light bulbs are still working at \( t = 1 \), we will use the given survival function \( P[X > t] = \frac{2}{2 + t} \) for \( t > 0 \).

First, let's denote the lifetimes of the two light bulbs by \( X_1 \) and \( X_2 \).

Given that both light bulbs are still working at \( t = 1 \), we need to determine the conditional probability that at least one light bulb is still working at \( t = 9 \).

### Step-by-Step Solution

1. **Survival Function Interpretation**:
   The survival function \( P[X > t] = \frac{2}{2 + t} \) indicates the probability that a light bulb lasts longer than \( t \) units of time.

2. **Conditional Probability for Individual Bulbs**:
   Given the survival function \( P[X > t] = \frac{2}{2 + t} \), we need to find the probability \( P[X > 9 \mid X > 1] \).

   By the definition of conditional probability:
   \[
   P[X > 9 \mid X > 1] = \frac{P[X > 9 \text{ and } X > 1]}{P[X > 1]}
   \]
   Since \( P[X > 9 \text{ and } X > 1] = P[X > 9] \) (because \( 9 > 1 \)):
   \[
   P[X > 9 \mid X > 1] = \frac{P[X > 9]}{P[X > 1]}
   \]

3. **Evaluating the Survival Function**:
   \[
   P[X > 9] = \frac{2}{2 + 9} = \frac{2}{11}
   \]
   \[
   P[X > 1] = \frac{2}{2 + 1} = \frac{2}{3}
   \]

4. **Calculating the Conditional Probability**:
   \[
   P[X > 9 \mid X > 1] = \frac{\frac{2}{11}}{\frac{2}{3}} = \frac{2/11}{2/3} = \frac{2}{11} \cdot \frac{3}{2} = \frac{3}{11}
   \]

5. **Probability that at Least One Bulb is Working at \( t = 9 \)**:
   We need to find the probability that at least one of the two bulbs is still working at \( t = 9 \), given both are working at \( t = 1 \).

   Let \( A \) and \( B \) be the events that bulb 1 and bulb 2, respectively, are working at \( t = 9 \) given they are both working at \( t = 1 \).
   \[
   P(A) = P(B) = \frac{3}{11}
   \]

   The probability that both bulbs are not working at \( t = 9 \) is:
   \[
   P(A^c \cap B^c) = P(A^c)P(B^c) = \left(1 - \frac{3}{11}\right)\left(1 - \frac{3}{11}\right) = \left(\frac{8}{11}\right)\left(\frac{8}{11}\right) = \frac{64}{121}
   \]

   Therefore, the probability that at least one bulb is working at \( t = 9 \) is:
   \[
   P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - \frac{64}{121} = \frac{57}{121}
   \]

### Conclusion
The conditional probability that at least one light bulb is still working at time \( t = 9 \), given that both light bulbs are still working at time \( t = 1 \), is \( \frac{57}{121} \).

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Would you like more details or have any questions about this solution? 

Here are some related questions you might find interesting:

1. How do you derive the survival function for different distributions?
2. What are the key differences between the survival function and the probability density function?
3. How would the problem change if the light bulbs had different survival functions?
4. What is the memoryless property in the context of exponential distributions?
5. How do you calculate joint survival functions for multiple random variables?
6. What is the significance of the conditional probability in reliability engineering?
7. How can Markov processes be applied to model the lifetime of components?
8. What are some practical applications of survival analysis in other fields?

**Tip**: Always check if the random variables are independent when dealing with joint probabilities. This can simplify your calculations significantly.

Explore the conditional probability of at least one light bulb still working at time t = 9, given that both are working at t = 1. Understand the application of survival functions and conditional probability in this advanced probability theory problem.

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