Type $i$ light bulbs function for a random amount
of time having mean $\mu_i$ and standard deviation
$\sigma_i, i = 1, 2$. A light bulb randomly chosen from a
bin of bulbs is a type $1$ bulb with probability $p$ and
a type $2$ bulb with probability $1 − p$. Let $X$ denote
the lifetime of this bulb. Find $E[X]$.
Just want to check if solution is correct:
Let $Y$ be indicator variable where $Y=1$ represents picking a Type 1 bulb and $Y=0$ represents picking a Type 2 bulb.
$E[X|Y=1]=\mu_1$
$E[X|Y=0] = \mu_2$
$E[X]=E[X|Y=1]P(Y=1) + E[X|Y=0]P(Y=0) = \mu_1p + \mu_2(1-p)$
Best Answer
Apart from two minor points, your solution is correct:
You use $T=1$ and $Y=1$ at different times to mean the same thing
You use $0$ and $2$ at different times to mean the not-$1$ type
So you perhaps should have written
$E[X|Y=1]=\mu_1$
$E[X|Y=2] = \mu_2$
$E[X]=E[X|Y=1]P(Y=1) + E[X|Y=1]P(Y=1) = \mu_1p + \mu_2(1-p)$