If bulbs fail at a time $T$ randomly in time, where they are exponentially distributed with parameter $\lambda \gt 0$. A room is lit with two bulbs with two bulbs with parameters $\lambda_1, \lambda_2$. How would I find the random variable that records the time at which the room becomes dark.
So far with one bulb I have $E(\lambda_1)=\frac{1}{\lambda_1}$ then similar for $E(\lambda_2)=\frac{1}{\lambda_2}$ , then since they are independent then the intersection (so the room going dark) is: $\frac{1}{\lambda_1 \lambda_2}$
Is this the correct way of thinking?
This is not the correct way of thinking. Let $X_1$ be a random variable denoting the time at which the first lightbulb fails, and let $X_2$ be a random variable denoting the time at which the second lightbulb fails. Under this notation, the time at which the room becomes dark is given by $\max(X_1, X_2)$. This is true because the room will be lit so long as one lightbulb is still working. So, you want to compute $\mathbb{E}[\max(X_1, X_2)]$.
There is a clever way to do this. Note that we can write
$$\mathbb{E}[X_1 + X_2] = \mathbb{E}[\max(X_{1}, X_{1}) + \min(X_{1}, X_{2})]. $$
We can rearrange and use linearity to find $\mathbb{E}[\max(X_{1}, X_{2})] = \mathbb{E}[X_1] + \mathbb{E}[X_{2}] - \mathbb{E}[\min(X_{1}, X_{2})]$. Also, we know that $\mathbb{E}[X_{1}] + \mathbb{E}[X_{2}]$ is just $\frac{1}{\lambda_1} + \frac{1}{\lambda_2}$.
All that remains is for us to calculate $\mathbb{E}[\min(X_1, X_2)]$. This can be done by noting that
$$F_{\min(X_{1}, X_{2})}(t) = P(\min(X_{1}, X_{2}) \leq t) = P(X_{1} \leq t, X_{2} \leq t) = P(X_{1} \leq t)P(X_{2} \leq t) = \lambda_1\lambda_2e^{-(\lambda_1 + \lambda_2)t}.$$
In other words, the random variable $\min(X_1, X_2)$ is exponentially distributed with parameter $\lambda_1 + \lambda_2$. This means that $\mathbb{E}[\min(X_{1}, X_{2})] = \frac{1}{\lambda_1 + \lambda_2}$, from which we conclude that the expected time until the room is dark is given by
$$\boxed{\mathbb{E}[\max(X_{1}, X_{2})] = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} - \frac{1}{\lambda_1 + \lambda_2}}$$