Lost Diamond Ring Probability Question Laura’s diamond ring is missing and is presumed to have equal probability of being 
accidently dropped in any of three shops she visited earlier in the day. If the ring is 
actually in shop 1 then the probability that the ring will be found upon a search of 
shop 1 is 0.23. If the ring is actually in shop 2 then the probability that the ring will be 
found upon a search of shop 2 is 0.51. If the ring is actually in shop 3 then the 
probability that the ring will be found upon a search of shop 3 is 0.26. What is the 
probability that the ring is in shop 1 given that the search of shop 1 was unsuccessful?
So my train of thought is it'll either .333 x 0.74 or just .333 because where it is is independent of searching for it.
 A: Hint:
Suppose Laura dropped $300$ rings ($100$ in each shop), and then shop 1 was searched


*

*How many rings would you expect to be found in shop 1?

*How many rings would you expect are still in shop 1 not found?

*How many rings would you expect to have not been found in total (shops 2 and 3 have not been searched)?

*Of the rings not found in total, what proportion do you expect to still be in shop 1?  

A: I would try it to solve using Bayes' theorem. First some notion
$R_i$: Ring was dropped in shop $i$
$S_i$: Search in shop $i$ was successful.
Bar means the complementary case: $\bar{S_1}$ = search in shop 1 was unsuccessful.
That means we have following identities:
$$
P(S_1|R_1) = 0.23, P(S_2|R_2) = 0.56, P(S_3|R_3) = 0.21.
$$
We would like to calculate $P(R_1|\bar{S_1})$ using
$$
P(R_1|\bar{S_1}) = \frac{P(\bar{S_1}|R_1)P(R_1)}{P(\bar{S_1})} = \frac{P(\bar{S_1}|R_1)P(R_1)}{P(\bar{S_1}|R_1)\cdot P(R_1) + P(\bar{S_1}|\bar{R_1})\cdot P(\bar{R_1})} = \frac{(1-0.23)\cdot0.33}{(1-0.23)\cdot0.33 + 1\cdot 0.66} = 0.278
$$
EDIT: The same answer you would get following advice of Henry :)
