Lifetime Probability and data loss I have an HDD and I know that the probability of failure of the HDD on any given day is p. I know that the expected lifetime of the HDD should be 1/p. What if I buy another HDD and copy everything inside? How can I find the expected time when I will lose some data (i.e. when both HDD break considering i'm not changing the first that fails)? 
I know that both have an expected lifetime of 1/p. Does it mean that they should break the same day or that I have less probability that both will break in the expected time that is 1/p?
 A: After reading over the comments it seems you have two components working in parallel (trying to keep the mathematical description of the problem general, so I use "components" instead of HDD, sorry). This means if one component fails the system still functions until the last component fails. So, the lifetime of the entire system depends on the lifetime of the last functioning component. If they all start in parallel at the same time, then it is the maximum lifetime time of the components we are concerned with finding the expectation (contrast with a series system where the failure of any one component shuts down the entire system so the failure time of the system is the minimum of all the component lifetimes).
Assume the component lifetimes are IID geometric RVs, $W_1, W_2$ with parameter $p$ taking values in $\mathbb{N}$, so that $\mathbb{E}(W_1)=1/p=\mathbb{E}(W_2)$. Assuming they are in use from the same starting time, and the system works until both components fail, then the entire system fails after $W=\max\{W_1, W_2\}$ time.
Note that $\{W\leq k\}=\{W_1 \leq k\} \cap \{W_2 \leq k\}$ so since these are IID, $\mathbb{P}(W\leq k)=\mathbb{P}(W_1 \leq k)^2$ follows. The CDF of a geometric RV with specified parameterization is $\mathbb{P}(W_1 \leq k)=1-q^k$ for $k=1,2,\dotsc$, where $q=1-p$. 
Then, the expected time until the entire system fails is 
$$\mathbb{E}(W)=\sum_{k=1}^\infty \mathbb{P}(W>k)=\sum_{k=1}^\infty 1-\mathbb{P}(W\leq k)$$
$$=\sum_{k=1}^\infty 1-\mathbb{P}(W_1\leq k)^2$$
$$=\sum_{k=1}^\infty 1-(1-q^k)^2$$
$$=\frac{q(q+2)}{(1-q)(1+q)}$$
$$=\frac{(1-p)(3-p)}{p(2-p)},$$
which then yields $E(W)\approx 1.667$ for $p=0.5$.
