1. ## 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?

2. 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:

$\Pr(\text{ Second shooter} | \text{ Hit}) = \frac{\frac{10}{19} \cdot \frac{7}{10}}{\frac{9}{19} \cdot \frac{8}{10} + \frac{10}{19} \cdot \frac{7}{10}} = ....$

3. I'm a bit confused.

Bayes' Theorem is:

$Pr(a|b)=\frac{Pr(b|a)*Pr(b)}{((Pr(b|a)*Pr(a))+((Pr (b|a^c)*(Pr(a^c))}$

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

4. 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

5. Originally Posted by downthesun01
I'm a bit confused.

Bayes' Theorem is:

$Pr(a|b)=\frac{Pr(b|a)*Pr(b)}{((Pr(b|a)*Pr(a))+((Pr (b|a^c)*(Pr(a^c))}$

How did you come to find that Pr(b|a) is equal to 10/19?
Did you draw the tree diagram?

6. 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

$\frac{(number\ of\ A\ hits)}{(number\ of\ A\ hits)+(number\ of\ B\ hits)}=\frac{70}{70+72}$ for the 2nd shooter.