Jointly Distributed Random Variables I have a question below I am stuck on. 
Suppose that you have ten lightbulbs, that the
lifetime of each is independent of all the other
lifetimes, and that each lifetime has an exponential
distribution with parameter $\lambda$.
a. What is the probability that all ten bulbs fail
before time $t$?
b. What is the probability that exactly $k$ of the ten
bulbs fail before time $t$?
c. Suppose that nine of the bulbs have lifetimes
that are exponentially distributed with parameter
$\lambda$ and that the remaining bulb has a lifetime
that is exponentially distributed with parameter
$\theta$ (it is made by another manufacturer). What is
the probability that exactly five of the ten bulbs
fail before time $t$?
For part a) I got that because the bulbs are independent, the answer is $ (1-e^{-\lambda t})^{10}$. I am currently stuck on b) and c) but I think that the answer to b) has to include $ (1-e^{-\lambda t})^{k}$ and $ (e^{-\lambda t})^{k}$ 
 A: Essentially you need to use the binomial distribution.  
For b) $p=1-e^{-\lambda t}$, so the probability that exactly $k$ bulbs fail is$\binom{10}{k}p^k(1-p)^{10-k}$.  
For c), you divide the probability into two parts (c1) the possibility that the odd bulb is one of the failures and (c2) the possibility that the odd bulb is not one of the failures.  Since these are mutually exclusive, the probability you need is the sum of the two.  Let $q=1-e^{-\theta t}$ the probability that the odd bulb fails.  Then c1) is given by $q\binom{9}{4}p^4(1-p)^5$ while c2) is given by $(1-q)\binom{9}{5}p^5(1-p)^4$.  Add these probabilities to get c).  To simplify arithmetic note that $\binom{9}{4}=\binom{9}{5}$.
A: 
For part a) I got that because the bulbs are independent, the answer is $ (1-e^{-\lambda t})^{10}$. 

Indeed.


I am currently stuck on b) and c) but I think that the answer to b) has to include $ (1-e^{-\lambda t})^{k}$ and $ (e^{-\lambda t})^{k}$ 

Yes, it almost does.   You want the probability that a selection of $k$ from $10$ bulbs fail before $t$ and that the remainder do not.  

 $$\binom {10}k\cdot (1-e^{-\lambda})^k\cdot(e^{-\lambda})^{(10-k)}\cdot \mathbf 1_{0\leq k\leq 10}$$


Now for (c), a hint: partition on whether the five failures include the $\theta$-rate bulb, and use the Law of Total Probability. Let $N_\lambda$ count the amount of failures among the nine $\lambda$-rate bulbs, and $N_\theta$ the failures among the one $\theta$-rate bulb. You seek:
$$\mathsf P(N_\lambda=5, N_\theta=0)+\mathsf P(N_\lambda=4, N_\theta=1)$$
