What is the Probability of Hitting the Target in a Two-Shooter Scenario?

[Physicsissuef]

Homework Statement

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: _______________

Homework Equations

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.

 

[Dick]

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?

 

[Physicsissuef]

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.

 

[Dick]

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?

 

[Physicsissuef]

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!

 

[HallsofIvy]

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?

 

[Physicsissuef]

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% ?

 

[Redbelly98]

82% is correct.

 

[Physicsissuef]

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

 

[Physicsissuef]

Some help please?

 

[fundoo]

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 %.

 

[Physicsissuef]

Where this formula comes from?

 

[fundoo]

if you are asking for

P(AnB) = P(A)xP(B)

then read Here ::

http://unfairstudy.uni.cc/4/41/591/2373/index.html

Under the Section "Independent Probabilities".

 

[Physicsissuef]

Ok, thank you.