Here's a pretty simple problem:
A factory produces 3 types of lamps, A, B, C. 30% of the lamps are of type A, 30% B and the rest 40% C. The lifetime of each type is distributed exponentially, such that $\lambda_A$ for A, $\lambda_B$ for B and $\lambda_C$ for C.
A lamp is considered to be valid if it's lifetime is more then $T$
A lamp is randomly selected. What is the probability that it is valid?
What I thought is the following:
Let X be the selected lamp, and $T_Y, Y\in \{A,B,C\}$ be the RV that represents the lifespan of a lamp. Therefore:
$P($X is valid$)=P(X=A, T_A\geq T) + P(X=B, T_B\geq T) + P(X=C, T_C\geq T)$
Now I am confused. For example, if I look on one of the above events, for instance:$P(X=A, T_A\geq T)$
How can I compute it? I mean, how can I for instance tell if the events $\{X=A\}$ and $\{T_A\geq T\}$ without any extra information? if they're not independent, how can I compute that?
The formula:
$P($ x is valid $ ) = P(X=A, T_A\geq T) + P(X=B, T_B\geq T) + P(X=C, T_C\geq T)$
is not exactly what you are looking for.
The correct formula would be :
P(x is valid) = P(A was chosen) P(x is valid | A was chosen) + P(B was chosen) P(x is valid | B was chosen) + P(C was chosen) P(x is valid | C was chosen)
Which gives:
$P( $ x is valid $ ) = \frac{3}{10} p(T_A \geq T) + \frac{3}{10} P(T_B \geq T ) + \frac{4}{10} P(T_c \geq T )$
This is because, if you know which lamp was chosen, you know exactly what the probability is that the lamp is valid, in terms of $\lambda$, because you know the way it is distributed.
The events "lamp A was chosen" and $T_A \sim exp(\lambda_A) $ are essentially the same.