Exponential distribution: Finding the parameter Please help me solve the following problem

Time of production of one electronic component is given with
exponential distribution with parameter λ. If the process lasts less than 3 hours, the component is working, otherwise it is defective. Events that component is working or that is defective have equally probability. Find:

*

*Parameter  λ

*Probability that the process will last less than 1 hour.


 A: Let $X$ be the random variable that measures length of life, in hours. We are told that $X$ has density function $\lambda e^{-\lambda x}$ (for $x\gt 0)$.
We are also told that $\Pr(X\lt 3)=\Pr(X\ge 3)$. Each is therefore $\frac{1}{2}$.
But $\Pr(X\gt x)$, by integration, is equal to $e^{-\lambda x}$. It follows that
$$e^{-3\lambda}=\frac{1}{2}.$$
Now we can solve for $\lambda$, by taking logarithm to the base $e$ of both sides.
And now that we have $\lambda$, we can find the probability the thing lasts less than $1$ hour, since $\Pr(X\le x)=1-e^{-\lambda x}$ for $x\gt 0$. 
Remarks: $1.$ We were asked to find $\lambda$, so we did. But note that the second problem can be solved without finding $\lambda$ explicitly. For we had $e^{-3\lambda}=\frac{1}{2}$. Taking cube roots, we find that $e^{-\lambda}=2^{-1/3}$, and therefore $\Pr(X\lt 1)=1-2^{-1/3}$. 
$2.$ The calculations of the problem  are closely related to half-life calculations that you have undoubtedly done several times in the past.  Individual atoms of a radioactive isotope have a lifetime that is exponentially distributed. The median lifetime is called the half-life of the substance. 
