Thread: Probability - shooting

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?

Originally Posted by losm1

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?

From a tree diagram and Bayes Theorem:

I'm a bit confused.

Bayes' Theorem is:



How did you come to find that Pr(b|a) is equal to 10/19?

I'm a bit confused.

Bayes' Theorem is:



How did you come to find that Pr(b|a) is equal to 10/19?

there is a hit

Rounds fired by both the guys = 19

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

Originally Posted by downthesun01

I'm a bit confused.

Bayes' Theorem is:



How did you come to find that Pr(b|a) is equal to 10/19?

Did you draw the tree diagram?

Originally Posted by losm1

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?

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.