Value of $x$ in a failure rate problem An electronic component is known to have a useful life which follows a probability
distribution with failure rate of $10^{-5}$
failures per hour. What is the probability that the
component would fail before the mean life of expected life?
This is Poisson Distribution right? Here the mean is $10^{-5}$ right but what is the value of $x$?
 A: A Poisson distribution is a counting distribution; it counts the random number of failures occurring in a fixed period of time.  Conversely, an exponential distribution is a failure time distribution; it measures the lifetime (time until failure) of a single component.
As such, a component that has a failure rate of $10^{-5}$ failures per hour translates (under the exponential lifetime model) to what average lifetime?  Think of it this way.  If on average we observe $10^{-5}$ failures per hour, then in $10^5$ hours, we would expect to observe $1$ failure, so the average lifetime of one component is $10^5$ hours.
Now, given this, what is the probability that the component happens to fail before its expected lifetime?  In other words, if $$X \sim \operatorname{Exponential}(\mu = 10^5)$$ where $\mu = \operatorname{E}[X] = 10^5$ hours, what is the probability $$\Pr[X \le \mu]?$$
A: @Heropup (+1) has already given you the correct distribution.
I think you're dealing with an exponential distribution with failure rate $\lambda,$ mean $\mu = 1/\lambda.$ and
CDF $F_X(x) = 1 - e^{\lambda x}.$ for $x > 0.$
So
$$P(X \le \mu) = 1 - e^{-\lambda\mu} = 1 - e^{-1} =  0.6321206.$$
In R, where pexp is a CDF:
lam = 10;  pexp(1/lam, lam)
[1] 0.6321206
lam = 1; pexp(1/lam, lam)
[1] 0.6321206
lam = .02; pexp(1/lam, lam)
[1] 0.6321206
lam = 10^-5; pexp(1/lam, lam)
[1] 0.6321206
1 - exp(-1)
[1] 0.6321206

Graph for one example. The relevant area is under the density curve to
the left of the vertical dotted line.

