Bayes Theorem

How would you use bayes theorem to calculate the probabilty of archer B hitting a target, given that one of them has.

Probabilities for the 3 archers:

A=0.6
B=0.3
C=0.5

I don't get it.  Why isn't the probability just 0.3?  The other results shouldn't affect B's probability.

Okay maybe it should be rephrased.

"One person has hit the target. What is the probability of that person being archer B".

The answer isnt just 0.3, that i am sure of.

without using bayes theorem i can get an estimate by doing 0.3 divided by the total. Which comes out at about 0.21 (done from memory)

say P(X) is the probability of player X hitting, n is the total number of hits

P(B|n=1) = P(n=1|B)*P(B)/P(n=1) = 1*0.3/(1-0.4*0.7*0.5) = 0.3488

dont trust this, i'm too lazy to tex and too lazy to check for mistakes

Out of 6+3+5 = 14 hits, 3 of them will be by B.  Therfoer >>4

>>4
This is just an intuitive statement of Bayes' Theorem in this context.

For events X and Y, we define P(X|Y) to be the probability of X occurring, given that Y has occurred.

Bayes': P(X|Y) = [P(Y|X)P(X)]/P(Y)

Let X be the event that B hits the target, and Y be the event that one of them hits the target.

P(X) = 0.3 as given
P(Y) = 0.6 + 0.3 + 0.5 = 1.4
P(Y|X) = P(one has hit, given that B has hit) = 1

So P(X|Y) = 0.3/1.4 = 0.214...

>>7
>> P(Y) = 0.6 + 0.3 + 0.5 = 1.4
P(Y) = 140%, you dont see anything wrong with this?