Please help me understand this probability statement.

Question

I was trying to answer this question:

A jewelry store that serves just one customer at a time is concerned about the safety of its isolated customers. The store does some research and learns that:

• 10% of the times that a jewelry store is robbed, a customer is in the store
• A jewelry store has a customer on average 20% of each 24-hour day
• The probability that a jewelry store is being robbed (anywhere in the world) is 1 in 2 million

What is the probability that a robbery will occur while a customer is in the store?

I have actually managed to answer the question. The answer is 1 in 4 million. However, what initially frustrates me is the semantics of the first information that we learnt. Which is the 10% of the times that a jewelry store is robbed, a customer is in the store.

When I read this, I could easily translates this to two meanings:

a) P(customer in the store | robbery occurring) = 0.10

or

b) P(robbery to happen | customer is in the store) = 0.10

The semantics here is confusing me. Even the final question that it is asking is kind of the same with the first statement that it is giving. Is there any way to differentiate the two?

Answer 1

The wording used is "10% of the times that a jewelry store is robbed, a customer is in the store". This means that if there are $N$ robberies, in $N/10$ of them there is a customer. That is, given that there is a robbery, the chance of there being a customer is 10%. So the correct interpretation is $P(Customer|Robbery)=1/10$.

The alternate interpretation $P(R|C)=1/10$ would mean that out of $N$ times that there is a customer in the store, in $N/10$ of them there is a robbery. That would be spoken of as "In 10% of the times that there is a customer, the store is robbed."

Given also $P(Customer)=2/10$ and $P(Robbery)=1/(2M)$, then $$P(R|C)=\frac{P(C\cap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}=\frac{\frac{1}{10}\frac{1}{2M}}{\frac{2}{10}}=\frac{1}{4M}$$

Answer 2

choice (a) is correct. it translates to:

the chance that a customer is in the store, given that a robbery is occurring. this is what the following translates to.

10% of the times that a jewelry store is robbed, a customer is in the store

choice (b) which is not correct, would translate to

chance that a store is robbed, given that a customer is in the store.

Given two events, A and B,

the rule is $p(A|B) \times p(B) = p(AB) = p(B|A) \times p(A).$

Thus, $p(A|B) = \frac{p(AB)}{p(B)}$,
while $p(B|A) = \frac{p(AB)}{p(A)}.$

Answer 3

I read the sentence as your interpretation a). We consider all situations where robberies happened so robbery is a precondition. And in 10% of these cases we had a customer in the store. So $P(\text{customer}| \text{robbery})=\frac{1}{10}$.

The situation b) could be expressed as "10% of the times that a customer was in the store, a robbery occurred". So it is possible to distinguish them.

The other givens are $P(\text{robbery}) = \frac{2}{1000000}$ and $P(\text{customer})=\frac{1}{5}$ (on any given moment).

Denoting the events customer present by $C$ and robbery occuring by $R$ for brevity, the text asks for $P(R|C)$ with $P(C|R), P(C), P(R)$ being given so this is a dead giveaway that Bayes' rule should be applied.