Question regarding Binomial Distribution Problem

Question

I am helping out a friend with some problems but I was a bit stuck on this one:

In a light-bulb factory, sample of 9 light-bulbs are taken from the production line every day to get tested. If all 9 are working, the sample passes the test. The probability that a light-bulb is broken is $p$. If 4 or more light-bulbs are broken, the sample simply fails the test. Moreover, if exactly one light-bulb is broken, a second sample of 6 light-bulbs is now drawn from the production line. If all those 6 light-bulbs are working, then the overall test is passed.

1. What is the probability of passing the test?

2. If $t$ represents the total number of light-bulbs removed from production line each day, (to conduct the tests), then:

   • Find the probability distribution of $t$ in terms of q = 1-p.

   • What is the expected number of light-bulbs taken away each day for testing, if $p = 0.05$?

Is it correct to assume that the discrete random variable $X: = "\text{sample light-bulb is broken}"$ follows a binomial distribution? If yes, is it correct to say that the probability of passing the test would be $$P(9W \cap \cup (1B \cap 6W)) = P(6W) + P(1B \cap 6W)$$ for $W: = \text{"The bulb is working"}$ and $B:= \text{"The bulb is broken"}$. Then, this translates to:

\begin{equation} P(9W \cap \cup (1B \cap 6W)) = {9\choose 0}q^9 + \left( {9\choose 1} p q^8 \right) \left( {6\choose 0} q^6\right) \end{equation}

Also, I am not sure how to go with 2).

Answer 1

1. Let $X_1 \sim Bin(9, p)$ and $X_2 \sim Bin(6,q)$. We are interested in $X_1=0$ or $(X_1=1,X_2=0) $. I can't just say that it is binomial distribution.

2a. Under your assumption in your comment, there are two ways to pass the test. Either to pass it with the first $9$ lightbulbs, or failed the first time and cleared the test with additional $6$ lightbulbs. No matter what you do, you will remove at least $9$ lightbulbs and another $6$ if the first batch has exactly one failure.

Hence just simplify $$P(t=9)=1-P(X_1=1)$$ $$P(t=15)=P(X_1=1)$$