I am trying to solve the following with Bayes Theorem but feel as if I am missing information:
Suppose you have a batch of red and a second batch of green
party-poppers. You know that one batch is reliable and will go off
with a bang 80% of the time whereas the other batch got wet and will
only explode 40% of the time. You choose a green party-popper at
random and pull the string. It doesn’t explode. Based on your first
observation what is the probability that the green batch is the
unreliable batch.
From the information provided I understand that there is a 50/50 chance of selecting either a red or green party popper. So, our prior must be 0.5. I also know that 80% of one batch explode, meaning 20% of that batch are unreliable and that 40% of the other batch explode meaning that 60% of that batch are unreliable. However, I do not understand how to plug in these values to Bayes Theorem in order to demonstrate the probability that the green batch is the unreliable batch as I do not know whether it is the green or red batch that is more likely to explode.
Any help with this query is very much appreciated.
Best Answer
The question is, given that a randomly picked green party popper did not explode, what is the probability of the green batch being unreliable. The first step to solving this is to translate the text into statements. As you noted, the prior is both batched being equally likely to be the unreliable one: $$ \Pr(\text{Green unreliable} ) = \Pr(\text{Red unreliable} ) = .5$$ The other two statements can be translated as follows (where $g=0$ indicates the event where the randomly selected popper does not explode) : $$ \Pr( g=0\ |\ \text{Green unreliable}) = 0.6 \\ \Pr( g=0\ |\ \text{Red unreliable}) = 0.2 $$
Then we can fill in these numbers in Bayes' theorem: \begin{align} \Pr( \text{G unreliable}\ |\ g=0) &= \frac{\Pr( g=0\ |\ \text{G unreliable}) \Pr( \text{G unreliable})}{\Pr( g=0\ |\ \text{G unreliable}) \Pr( \text{G unreliable}) + \Pr( g=0\ |\ \text{R unreliable}) \Pr( \text{R unreliable})}\\ &= \frac{0.6}{0.6+0.2}\\ &= \frac{3}{4} \end{align}