Bayes Theorem with Missing Information

Question

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.

Answer 1

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}

Answer 2

I think you can just leave the (confusing) colors out and rephrases like this:

"You choose a party-popper at random out of one of the batches (fifty-fifty) and pull the string. It doesn’t explode. Based on your first observation what is the probability that this party-popper comes from the unreliable batch?"

For this you can use Bayes rule:

$P\left(\text{unreliable}\mid\text{no explosion}\right)=\frac{P\left(\text{no explosion}\mid\text{unreliable}\right)P\left(\text{unreliable}\right)}{P\left(\text{no explosion}\mid\text{unreliable}\right)P\left(\text{unreliable}\right)+P\left(\text{no explosion}\mid\text{reliable}\right)P\left(\text{reliable}\right)}$

Answer 3

I do declare that we must assume the batches are of the same size, there is no bias as to which batch was wet, and that wetness does not bias from which batch the popper was selected.

We shall designate the relevant events as being that you select a popper which: is green as $E$, is unreliable as $U$, and does explode as $E$.

The above assumptions and the data provided indicates the following probability measures: $$\begin{align}\mathsf P(G)&=1/2\\\mathsf P(U\mid G)&=1/2\\ \mathsf P(E\mid U)&=2/5&&= \mathsf P(E\mid U,G)\tag{ *} \\\mathsf P(E\mid U^\complement)&=4/5&&=\mathsf P(E\mid U^\complement,G)\tag{*}\end{align}$$

($*$ by reasoning that Explosivity is determined only by the Unreliability of the batch, and not (directly) by its Colour, therefore Colour and Explosivity are independent for a given state of Unreliablity ($E\perp G\mid U$)).

Now the law of total probability states:

$$ \mathsf P(E\mid G)=\mathsf P(E\mid U,G)\mathsf P(U\mid G)+\mathsf P(E\mid U^\complement,G)\mathsf P(U^\complement\mid G) \\ \mathsf P(E^\complement\mid G)=\mathsf P(E^\complement\mid U,G)\mathsf P(U\mid G)+\mathsf P(E^\complement\mid U^\complement,G)\mathsf P(U^\complement\mid G) $$

You seek $\mathsf P(U\mid G,E^\complement)$. Use Bayes' Rule to find this.