Expected value of minimum of exponential random variables I'm working on the following question:
A device contains two components, A and B. The lifespans of A and B are both exponentially distributed with expected lifespans of 5 years and 10 years respectively. The device works as long as both components work. What is the expected lifespan of the device?
I'm aware that A ~ Exp(1/5) and B ~ Exp(1/10). initially I thought that knowing the expected value of both variables (10 and 5) would allow me to use the linearity of expectations:
E[A + B] = E[A] + E[B] = 15
Then I realised that would be the expected total years of life of both components, which isn't what we're looking for.
The answer given in the textbook is λA + λB = λ(3/10) . The expected value of this is 3.3 years.
Intuitively , this seems correct, even though I didn't know how to sum exponentials like that. What I'm trying to understand though is how my application of the linearity of expectations is not accurate. Why is E[A + B] = 3.3 when E[A] + E[B] = 15 ?

This is a problem where you want to examine the lifetime of a system composed of two components operating in series.  It is part of the broader subject of reliability analysis (also called survival analysis when looking at survival of organisms).  I recommend you read some introductory resources to get a general understanding of this subject area, including an understanding of the analysis of series and parallel systems.
In this particular case we have a system time-to-failure $T_\text{sys} = \min(A,B)$ and we want to find the expected system time-to-failure $\mathbb{E}(T_\text{sys})$.  To do this, we first need to derive the distribution of the system time-to-failure and then we use this to find the expected time-to-failure for the system.  Since the times-to-failure for the two components are independent exponential random variables we have:
$$\begin{align}
F_\text{sys}(t) 
&= \mathbb{P}(T_\text{sys} \leqslant t) \\[6pt]
&= \mathbb{P}(\min(A,B) \leqslant t) \\[6pt]
&= 1 - \mathbb{P}(\min(A,B) > t) \\[6pt]
&= 1 - \mathbb{P}(A > t, B > t) \\[6pt]
&= 1 - (1-F_A(t)) (1-F_B(t)) \\[6pt]
&= 1 - \exp(-\lambda_A t) \exp(-\lambda_B t) \\[6pt]
&= 1 - \exp(-(\lambda_A + \lambda_B) t), \\[6pt]
\end{align}$$
so we can see that $T_\text{sys} \sim \text{Exp}(\lambda_A + \lambda_B)$.  Now, taking its expected value we get:
$$\begin{align}
\mathbb{E}(T_\text{sys})
&= \int \limits_0^{\infty} (1-F_\text{sys}(t)) dt \\[6pt]
&= \int \limits_0^{\infty} \exp(-(\lambda_A + \lambda_B) t) dt \\[6pt]
&= \Bigg[ - \frac{1}{\lambda_A + \lambda_B} \cdot \exp(-(\lambda_A + \lambda_B) t) \Bigg]_{t=0}^{t \rightarrow \infty} \\[6pt]
&= \Bigg[ ( 0 ) - \bigg( - \frac{1}{\lambda_A + \lambda_B} \bigg) \Bigg] \\[6pt]
&= \frac{1}{\lambda_A + \lambda_B}. \\[6pt]
\end{align}$$
Substitution of your particular parameters gives you the result given by your textbook.  As you can see from the above working, when you have components in series and these components operate independently with exponential time-to-failure, the system time-to-failure is exponentially distributed with hazard rate equal to the sum of the hazard rates of the components.  This is a well-known result for series systems in reliability analysis.
