Probability, who is right?

Question

A thief who is escaping has three paths to choose from A, B and C. Path B splits into 2 paths, BA leading to path A and BC leading to path C. In A he cannot escape, in C he escapes. If it is known that the thief escaped, what is the probability that he used path BC? I have a problem with this question, my friend say me this:

$\frac{(0.33*0.5)}{(0.33*1)+(0.33*0.5)}=0.33$

But i think that doesnt make any sence, i think that the answer is this

$\frac{0.5}{0.5+0.5}=0.5$

this is because we have the condition that the thief escaped, it can be assumed that he can only go by C or BC

Who is right?

Answer 1

I think the following way helps to think about such problems.

Imagine there were 6 thieves. Two went to A, one to BA, one to BC, two to C. Out of this 3 survived, and from these three survivals there were just 1 who went through BC, so given that thief survived, probability of them going through BC is $1/3$.

Answer 2

Assuming selections of A , B , C is equally probable probability of doing so is $1/3 $
After choosing B again assuming splitting into BC , BA is equally likely probability of doing so is $1/2$ for each case .
Now probability of escaping is $$1/3 + (1/3)*(1/2) = 1/2 $$and probability of escaping by BC is $$(1/3)*(1/2) $$
Probability of using BC given that he has escaped is given by second answer divided by first answer because that is a conditional probability, so the answer is $1/3$.