Two guys are shooting at a target. First one fires 9 rounds and second one fires 10 rounds in this session. First one has 8/10 precision and second one has 7/10 precision. There is a hit. What is the probability that second shooter is the one that hit it?
Rounds fired by both the guys = 19there is a hit
Rounds fired by Second one = 10
Probability that the round that hit the target was fired by 2nd guy = 10/19
Probability that the second one hit the target = 7/10
This should occur simultaneously
......What's wrong with that ?I don't find anything wrong in MrF's reply
the 2nd shooter hits the target an average of 7 out of 10 times,
so it's expected he should hit the target 7 times from his 10 attempts
(or 70 times per hundred etc)
the first shooter hits the target an average of 8 times out of 10,
so he ought to hit the target an average of 9(0.8)=7.2 per 9,
or 72 times per ninety etc.
Then, the probability of a single hit belonging to one or the other is
for the 2nd shooter.