Beginning at time $t=0$, we start using bulbs, one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a type-A bulb and a type-B bulb. The lifetime $X$, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:
Type A : $ f_{X}(x,y)=\cases{e^{-x} & if $x \ge 0$\cr 0 & otherwise\cr} $
Type B : $ f_{X}(x,y)=\cases{3e^{-3x} & if $x \ge 0$\cr 0 & otherwise\cr} $
Determine the probability that the total period of illumination provided by the first two type-B bulbs is longer than that provided by the first type-A bulb.
Let $I_A,I_B$ be the illumination period provided by a single type A,B bulb.With other words $I_A,I_B$ are the lifetimes of our bulbs (described by the above PDF)
$$ P(2*I_A < t \space and \space 2*I_B > t) = P(I_A < t/2) *(1-P(I_B < t/2)) $$ $$ =(1-e^{-1* \frac t 2})(1-(1-e^{-3 * \frac t 2})) $$ Now clearly my "answer" depends on $t$ , how can I pursue my answer further and git rid of the $t$ and have an exact number as a result ? (if it's wrong, please point out why)
The answer is using an Erlang distribution and results to $\frac {7} {16}$
Hint. The sum of two exponential random variables is an erlang distribution. The total illumination period of the first two lightbulbs of type B is thus erlang distributed.
Using this, the question boils down to determining $\mathbb{P}(erlang(2,3) > exp(1))$
Edit. I noticed that I read the question wrong. I updated my answer to answer the question.