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
Looking at it logically, we get
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.