The time it takes for a candle having a lifetime that follows an exponential distribution to go off We have $5$ candles each having a lifetime which follows an exponential distribution with parameter $\lambda$. We light up each candle at time $t=0$.
Assume that $Y$ is the time that it takes for the third candle to go off. What is the expectation and variance of $Y$?
My try:
First of all, I believe having $5$ candles is irrelevant. We only need to consider one random variable following the exponential distribution, like $X\sim \exp(\lambda)$. It means that on average, it takes $\frac{1}{\lambda}$ for the candle to go off. However, this does not seem like a random variable. It seems like it is a constant. Then, it won't be meaningful to calculate the expectation and variance. Am I right?
Also, we know that at some point, the candle "will" go off. So, does this mean that we cannot predict at which time it will? I am totally confused thinking about these concepts. I appreciate if someone enlightens me.
Note: There is a similar question here. However, the question has not been answered due to the lack of attemps provided by the OP.

The time needed for the first candle to go out is distributed as the minimum of $5$ independent exponential random variables. That minimum is exponentially distributed with rate $5\lambda$ (see here). Since the exponential distribution is memoryless we now start all over; by similar reasoning as before the second-out and third-out times are exponentially distributed with rates $4\lambda$ and $3\lambda$ respectively, and all three new random variables are independent of each other. $Y$ is their sum; since the mean and variance of an exponential distribution with rate $\lambda$ are $\frac1\lambda$ and $\frac1{\lambda^2}$
$$\mathbb E[Y]=\frac1{5\lambda}+\frac1{4\lambda}+\frac1{3\lambda}=\frac{47}{60\lambda}$$
$$\operatorname{Var}(Y)=\frac1{25\lambda^2}+\frac1{16\lambda^2}+\frac1{9\lambda^2}=\frac{769}{3600\lambda^2}$$
