Let's solve the questions based on the image provided. Here's a step-by-step breakdown:

(a)

The problem involves two bulbs with a lifetime modeled by an exponential density function with a mean $\mu = 1500$ hours. We're asked to find the probability that both bulbs fail within $1800$ hours.

Step 1: Exponential distribution formula

For an exponential distribution with mean $\mu$, the rate parameter $\lambda$ is given by: $\lambda = \frac{1}{\mu} = \frac{1}{1500}.$ The cumulative distribution function (CDF) for an exponential distribution is: $P(T \leq t) = 1 - e^{-\lambda t}.$

Step 2: Probability for one bulb

The probability that one bulb fails within $t = 1800$ hours is: $P(T \leq 1800) = 1 - e^{-\lambda \cdot 1800}.$ Substitute $\lambda = \frac{1}{1500}$: $P(T \leq 1800) = 1 - e^{-\frac{1800}{1500}} = 1 - e^{-1.2}.$

Step 3: Probability for two independent bulbs

For two independent bulbs, the joint probability that both fail within $1800$ hours is: $P(\text{Both fail}) = P(T_1 \leq 1800) \cdot P(T_2 \leq 1800).$ Using the result for one bulb: $P(\text{Both fail}) = \left(1 - e^{-1.2}\right)^2.$

Let’s compute this numerically: $1 - e^{-1.2} \approx 1 - 0.3012 = 0.6988.$ $P(\text{Both fail}) = (0.6988)^2 \approx 0.4883.$

So, the probability that both bulbs fail within 1800 hours is approximately 0.4883.

(b)

Now, consider a single bulb from part (a). If it burns out, it is replaced by a new bulb of the same type. We need the probability that the two bulbs fail within a total of $1800$ hours.

Step 1: Explanation

In this case, the lifetime of the first bulb and the replacement bulb are independent exponential random variables with mean $1500$. The total time until both bulbs fail is the sum of two independent exponential random variables, which follows a Gamma distribution with shape parameter $k = 2$ and rate parameter $\lambda = \frac{1}{1500}$.

Step 2: CDF of a Gamma distribution

The probability that the total failure time is within $1800$ hours is: $P(T_{\text{total}} \leq 1800) = \int_0^{1800} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{(k-1)!} \, dt,$ where:

• $k = 2$,
• $\lambda = \frac{1}{1500}$.

Substitute these values: $P(T_{\text{total}} \leq 1800) = \int_0^{1800} \frac{\left(\frac{1}{1500}\right)^2 t e^{-\frac{t}{1500}}}{1} \, dt.$

Step 3: Solve using the Gamma distribution

The integral simplifies using the CDF of the Gamma distribution: $P(T_{\text{total}} \leq 1800) = 1 - \left(e^{-\frac{1800}{1500}} \left(1 + \frac{1800}{1500}\right)\right).$

Compute numerically: $P(T_{\text{total}} \leq 1800) \approx 1 - \left(0.3012 \cdot (1 + 1.2)\right) = 1 - \left(0.3012 \cdot 2.2\right).$ $P(T_{\text{total}} \leq 1800) \approx 1 - 0.6626 = 0.3374.$

Thus, the probability that the two bulbs fail within a total of $1800$ hours is approximately 0.3374.

Additional Questions

1. Can you explain why the Gamma distribution applies in part (b) instead of just multiplying probabilities?
2. How would the probabilities change if the mean lifetime of the bulbs were doubled to $3000$ hours?
3. What is the significance of the independence assumption in these calculations?
4. How would the calculation in part (b) differ if we had three sequential replacements instead of one?
5. Could we approximate part (b) using simulation instead of analytical methods? How?

Tip

For exponential distributions, the memoryless property states that the remaining lifetime of a bulb is independent of how long it has already been running—this is key to understanding problems like these!