What statistical test do I need? Say I have $N$ light bulbs. Whevener one breaks down, I immediately fix it. 
$k_0$ of these $N$ light bulbs do not break down during this year,
$k_1$ break down (and get fixed) once, and
$k_2$ break down (and get fixed) twice. 
My question is how to test whether $k_2$ is too large to what we would expect if the distribution is Poisson, that is, whether previously fixed light bulbs have a greater chance of breaking down again, and what the $p$-value is. 
Thank you in advance!
 A: One conventional option is a likelihood ratio test (LRT) between the Poisson model and a saturated model. "Saturated model" just means a model with no constraints on P(0 breakdowns), P(1 breakdown), and P(2 breakdowns) besides the fact that they're positive and they sum to 1. The LRT test statistic is $-2ln(\frac{L_0(\hat \theta_0)}{L_A(\hat \theta_A^*)})$, where $L$ is a likelihood, $\hat \theta$ is the parameter value maximizing that likelihood, and subscripts indicate the Poisson model ($0$) or the saturated model (A). In this case, Wilks' Theorem states that the distribution of this statistic is approximately $\chi^2_{1}$ (under the null), so you can get p-values or critical values from any chi-squared table.
RE: computation, the saturated max-likelihood parameter estimates are $p_0, p_1, p_2 = (k_0, k_1, k_2)/N$, with the multinomial likelihood ${N \choose k_0, k_1, k_2}p_0^{k_0}p_1^{k_1}p_2^{k_2}$. The Poisson max-likelihood estimator is $\hat \theta_0 = \frac{k_1 + 2k_2}{N}$, and you would plug that into the Poisson likelihood $L_0(\theta) = \prod_j (\frac{\exp(-\theta)\theta^j}{j!})^{k_j}$.
If you're feeling ambitious, you could use a version of the LRT where you only allow an excess of double breakdowns, and not a deficit. That might use theory like this.
A: Maybe I have a simple test for your problem.
My question is how to test...whether previously fixed light bulbs have a     
greater chance of breaking down again, and what the p-value is. 



*

*Model


Let's suppose you have a sample of $n$ light bulbs. For each $i$ light bulb define the random variables:
$$B_{1,i} = 
\left\{\begin{array}{l,l}
1, &\text{if the i-th light bulb broke the first time}\\
0, &\text{if it didn't broke the first time}
\end{array}\right.$$
$$B_{2,i} = 
\left\{\begin{array}{l,l}
1, &\text{if the i-th light bulb broke a second time}\\
0, &\text{if it didn't broke after been fixed}
\end{array}\right.$$
which we assume independent. Using this notation we have: $k_1 = \sum_{i = 1}^nB_{1,i}$ and $k_2 = \sum_{i = 1}^nB_{1,i}B_{2,i}$.
Lets denote the probability of a light bulb breaking the first time as $p_1$ and the probability of it braking a second time as $p_2$. Then the statistical hypothesis can be formulated as $H_0$: $p_1 = p_2$ and $H_1$: $p_1<p_2$.


*

*Test


I want to propose a test based on the statistic:
$$T = \frac{1}{n}\sum_{i = 1}^nB_{1,i}(1-B_{2,i}) = \frac{k_1 - k_2}{n}$$
To see how such a test would work notice that $\mathbb{E}(T)=p_1(1-p_2)$. Now if $p_1<p_2$ then $p_1(1-p_2)>p_1(1-p_1)$. So a test based on $T$ reject $H_0$ if $T$ is bigger than one would expect.
There are two ways to find the associated p-value. The hard way is to use the true distribution of $nT$ which is a binomial distribution with mean $p_1(1-p_1)$ (assuming $H_0$ is true). The easy way is to use the normal approximation:
$$\sqrt{n}\frac{T - v}{\sqrt{v(1-v)}}\overset{H_0}{\longrightarrow} N(0,1),$$
where $v = \frac{1}{n-1}\sum_{i=1}^{n}(B_{1,i} - \bar{B_{1}})^2$ and $\bar{B_{1}} = \frac{1}{n}\sum_{i=1}^nB_{1,i}$.
