# Probability: A Rifle Shooter

1. Jan 4, 2013

1. The problem statement, all variables and given/known data

A rifle shooter aims at a target, at a distance d, but has an accuracy controlled by a probability density:
$\rho(\phi)=\frac{1}{2\Phi}$

$\phi\in(-\Phi,\Phi)$​

where $\phi$ is the angle achieved and is bounded by the small angle $\Phi$.

(Refer to attachment for diagram)

Calculate the probability density for where the bullet strikes the target, $\tilde{p}(x)$. (I've done this.. I think)

If a target is set up with a width of 2d, with success H say, being hitting the target and failure M say, being missed the target, calculate and depict the probability of hitting,

$P(H;\Phi)$
as a function of $\Phi$ for fixed d and D with $D=tan(\theta)$
Hint: Be careful about when $\Phi=\phi$

2. Relevant equations

$p(a,b)=\int{dx p(x)}$

3. The attempt at a solution

For the bit I've done,
I got $\tilde{p}(x)=\frac{2\Phi}{\pi}$

I don't know where to start on the next bit, could someone give me a hint please?

File size:
3.6 KB
Views:
235
2. Jan 4, 2013

### Simon Bridge

For best results, you should show your reasoning for the bit you've done.
How did you work that out?
What does x represent in p(x)? How is it related to $\phi$?
What is D and d (and $\theta$)?

You have a probability density function - how would you normally go about calculating a probability from that?

3. Jan 4, 2013

For the bit I've done:
P(a, b)=$\int_a^b \! p(x) \, \mathrm{d} x.$
$P(x, x+\delta x)=\delta\theta\frac{1}{\pi}=p(x) \delta x$ (I got this from what our lecturer did in a question with a person throwing balls randomly toward a wall.)
so,
$p(x)=\frac{1}{\pi} \frac{d\theta}{d x}$

$now, tan \theta = \frac {x}{D}$

$so, \frac{d\theta}{dx}= \frac {D}{D^2+x^2}$

$so, p(x) = \frac{1}{\pi} \frac {D}{D^2+x^2}$

$then, \tilde{p}(x)=p(-D tan\Phi, D tan\Phi)$

$= 2 \int_0^{Dtan \Phi} \! \frac{1}{\pi} \frac {D}{D^2+x^2}$

$= \frac{2\Phi}{\pi}$

D is the distance from the shooter to the target. (shown in attached diagram)

Last edited: Jan 4, 2013
4. Jan 4, 2013

I would integrate it over the interval I want the probability for ?

5. Jan 4, 2013

### Simon Bridge

D = the distance to the target (you did that - good)
d = the radius of the target
x = radius to where a bullet went at the target distance - so that $$p(x,x+\delta x)=\frac{2\Phi\delta x}{\pi}$$... this is what you are telling me?