# Archery probability

• Mar 17th 2011, 01:26 PM
dwsmith
Archery probability
In archery, targets are marked with 10 evenly spaced concentric rings, which have values 1 through 10 assigned to them. In addition, there is an inner 10 ring, sometimes called the X ring. This becomes the 10 ring at indoor compound competitions. Outdoor it serves as a tie breaker, with the archer scoring the most Xs winning. Suppose we have an outdoor competition using an olympic size target, which has a diameter of 12cm. When a particular archer shoots an arrow, it is equally likely to fall anywhere on the target, but it will hit the target.
Find the probability function of the score obtained from the archer shooting one arrow at the target.

I tried binomial but this didn't work

$\displaystyle \displaystyle p(x)=\binom{10}{x}\left(\frac{1}{10}\right)^x\left (\frac{9}{10}\right)^{10-x}$
• Mar 17th 2011, 01:35 PM
pickslides
Hi dw, what does $\displaystyle \frac{1}{10}$ represent here?

If the probability is dependant on the area of each evenly spaced toroid, is each area the same?
• Mar 17th 2011, 01:39 PM
dwsmith
Quote:

Originally Posted by pickslides
Hi dw, what does $\displaystyle \frac{1}{10}$ represent here?

If the probability is dependant on the area of each evenly spaced toroid, is each area the same?

1/10 is the probability of hitting a ring. I don't know really know what to do for this problem.
• Mar 17th 2011, 01:53 PM
Plato
To be quite honest, I do not follow the makeup of the target.
But I do know that this is not binominal.
The probability of hitting inside any particular ring is the ratio of the area of that ring to the total of the target area.

So the probability of hitting the outside ring is its area divided by the total target area. This is where I don’t understand how the rings work.
• Mar 17th 2011, 04:59 PM
Soroban
Hello, dwsmith!

Quote:

In archery, targets are marked with 10 evenly spaced concentric rings,
which have values 1 through 10 assigned to them.

Suppose we have an olympic-size target, which has a diameter of 12cm. .??
When a particular archer shoots an arrow, it is equally likely to hit
anywhere on the target, but it will hit the target.

Find the probability function of the score obtained
from the archer shooting one arrow at the target.

That's a very small target . . . and it has ten rings, too!

The probability of a particular score is the ratio of the area of that ring
. . to the area of the entire target.

Suppose the target has radius $\displaystyle \,R.$ .The area of the target is: $\displaystyle \pi R^2$

The "10-circle" has a radius of $\displaystyle \frac{R}{10}$
. Its area is: $\displaystyle \pi(\frac{R}{10})^2 = \frac{\pi}{100}R^2$

The "9-circle" has a radius of $\displaystyle \frac{2R}{10}$
. Its area is: $\displaystyle \pi(\frac{2R}{10})^2 =\frac{4\pi}{100}R^2$
The "9-ring" has an area of: $\displaystyle \frac{4\pi}{100}R^2 - \frac{\pi}{100}R^2 \:=\:\frac{3\pi}{100}R^2$

The "8-circle" has a radius of $\displaystyle \frac{3R}{10}$
. Its area is: $\displaystyle \pi(\frac{3R}{10})^2 =\frac{9\pi}{100}R^2$
The "8-ring" has an area of: $\displaystyle \frac{9\pi}{100}R^2 - \frac{4\pi}{100}R^2 \:=\:\frac{5\pi}{100}R^2$

We see a pattern emerging . . . and we can complete our table.

. . $\displaystyle \begin{array}{ccc} \text{Ring} & \text{Area} & \text{Prob.} \\ \hline \\[-3mm] \text{1-ring} & \frac{19\pi}{100}R^2 & \frac{19}{100} \\ \\[-3mm] \text{2-ring} & \frac{17\pi}{100}R^2 & \frac{17}{100} \\ \\[-3mm] \text{3-ring} & \frac{15\pi}{100}R^2 & \frac{15}{100} \\ \\[-3mm] \text{4-ring} & \frac{13\pi}{100}R^2 & \frac{13}{100} \\ \\[-3mm] \text{5-ring} & \frac{11\pi}{100}R^2 & \frac{11}{100} \\ \\[-3mm] \text{6-ring} & \frac{9\pi}{100}R^2 & \frac{9}{100} \\ \\[-3mm] \text{7-ring} & \frac{7\pi}{100}R^2 & \frac{7}{100} \\ \\[-3mm] \text{8-ring} & \frac{5\pi}{100}R^2 & \frac{5}{100} \\ \\[-3mm] \text{9-ring} & \frac{3\pi}{100}R^2 & \frac{3}{100} \\ \\[-3mm] \text{10-ring} & \frac{\pi}{100}R^2 & \frac{1}{100} \\ \\[-3mm] \hline\end{array}$

If $\displaystyle \,n$ is the number of the ring desired,

. . then: .$\displaystyle P(n) \:=\:\dfrac{21 - 2n}{10}$