# Probability homework problem

1. Jul 19, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data
Two shooters aim in same target at same time and then the first shooter shots in 70% of the shooting session, and the second is missing in the 60% of the shooting sesion.

A. The probability to miss the first shooter is: ______________

B. The probability the target to be shot is: _______________

2. Relevant equations

3. The attempt at a solution

A. I think it is 30% (very easy indeed)

So the first shooter shots 70% (7 of 10) and the second shooter 40% (4 of 10)

B. I think the probability is more than 70%, but what is the correct one?

Thank you.

2. Jul 19, 2008

### Dick

Re: probability

If you found A easy then what's the probability BOTH shooters miss? What's the relation of that probability to the probability that the target is hit?

3. Jul 20, 2008

### Physicsissuef

Re: probability

The both shooters to miss is probably 55% (or 5,5 of 10)

$$1-\frac{5,5}{10}=\frac{4,5}{10}$$

I don't think this is correct.

4. Jul 20, 2008

### Dick

Re: probability

You're right. It's not. Odds A misses are 0.30, odds B misses are 0.60. Odd that they both miss is NOT the average of the two. What is it?

5. Jul 20, 2008

### Physicsissuef

Re: probability

If the first one misses 3 of 10, and the second one 6 of 10, both miss 9 of 20, or 45%. I don't really know. Please help!

6. Jul 20, 2008

### HallsofIvy

Staff Emeritus
Re: probability

Do you know how to find P(A and B) if you know P(A), P(B), and know that A and B are independent?

7. Jul 20, 2008

### Physicsissuef

Re: probability

Ok, I will do like this:
1-shot 2-miss

shooter C - 1111111222

shooter D - 1111222222

All possible combinations are:
7*(1,1 ; 1,1 ; 1,1 ; 1,1 ; 1,2 ; 1,2 ; 1,2 ; 1,2; 1,2; 1,2)+3*(2,1 ; 2,1; 2,1 ; 2,1 ; 2,2 ; 2,2; 2,2 ; 2,2 ;2,2 ;2,2)
10*7+3*10=70+30=100

$$\frac{10*7+4*3}{100}=\frac{70+12}{100}=\frac{82}{100}$$

82% ?

8. Jul 20, 2008

### Redbelly98

Staff Emeritus
Re: probability

82% is correct.

9. Jul 21, 2008

### Physicsissuef

Re: probability

Ok, thanks. But how will I solve it with permutations?

10. Jul 21, 2008

### Physicsissuef

Re: probability

11. Jul 21, 2008

### fundoo

Re: probability

for second question ::

probability that A miss :p(A): 30/100
probability that B miss :p(B): 60/100

Probability that both Miss :: p(AnB) = p(A) x p(B) = 30/100 * 60/100 = 18/100 (As both Events are independent)

so the Probability that the Target is shot = 1 - Both Miss
1 - 18/100 = 82/100 = 82 %.

12. Jul 21, 2008

### Physicsissuef

Re: probability

Where this formula comes from?

13. Jul 21, 2008

### fundoo

14. Jul 22, 2008

### Physicsissuef

Re: probability

Ok, thank you.