Question

http://www.ode.state.or.us/wma/teachlearn/subjects/mathematics/assessment/worksamples/h.2s.3_dartthrowing.doc Can some please explain how to solve this for me? It's a circle problem btw.

Answer 1

You should first find the total area of the circle, and then the area of each "section"

Answer 2

I know to find the area I need to use A=(3.14)r^2, right?

Answer 3

In theory, the probability of hitting a section is the area of the section divided by the total area of the circle.

so step 2: find the probability of hitting each of the sections

step 3: expected value (the score) is :
sum up each section: score for a section * prob of that section

multiply that value by 100 to get the expected value of 100 tosses

Answer 4

The four quadrants inside the small circle average to 7.5 points.
The areas inside the large circle outside the small circle average to 2.5 points each.
You can consider each area of these two (7.5 points and 2.5 points) without having to consider all eight areas separately.

Answer 5

I'm still so lost... Sorry guys I'm terrible at mathematics :(

Answer 6

Do you get the idea that the chance of hitting a section depends on how big the section is ?
i.e its area ?

Answer 7

Yes I understand that. Not sure it's area. how do I find the area? I know the formula but not sure how.

Answer 8

The idea is it's 100% sure you hit the whole circle and it's more likely to hit a "big target" than a small target. In the real world it's more complicated (people who have a good eye tend to get close to where they aim. We will ignore that, and just assume: big area, better chance of hitting it.

Look at the "inner circle". Do you know its radius ?

Answer 9

They tell you the inner circle has a diameter of 4, so its radius is ½ of that or 2
the area of the inner circle (using pi r^2 ) is pi*2*2 = 4pi

Answer 10

the entire target has diameter 8 meters, so radius 4 and area pi*4*4 = 16 pi

Answer 11

the chance of hitting the inner circle is the area of the small circle divided by the big circle
\[ \frac{4 \pi}{16 \pi}= \frac{1}{4} \]

Answer 12

we don't really want that , we want the chance of hitting one of the "sectors"
but based on the picture, each sector is ¼ of the inner circle, so each has area ¼ of the inner circle: ¼ of 4pi is pi
and the chance of hitting one of the sectors in the inner circle is
\[ \frac{\pi}{16\pi}= \frac{1}{16} \]

the chance of hitting any of the 4 sectors in the small circle is 1/16
make a note of that.

Answer 13

next, you want the area of the "outer ring"
that area is the area of the big circle minus the area of the inner circle
in other words, its area is
\[ 16\pi - 4\pi = 12 \pi\]

Answer 14

It looks like each of the sectors in the outer ring are ¼ of the ring
so each has area ¼ * 12 pi = 3 pi

and the chance of hitting any of these sectors is
\[ \frac{3 \pi}{16 \pi}= \frac{3}{16} \]
make a note of that, also

Answer 15

Once you have the probabilities, you find the "expected value"
which is the "value" of a sector times its probability
and you add them up for all the sectors (all 8)

Answer 16

The outer sectors are have probability 3/16
so we do
3/16*2 + 3/16*3 + 3/16 * 2 + 3/16* 3
which is 3/16*(2+2+3+3) or 3/16 * 10 = 15/8
and the expect value of the inner sectors (each with prob 1/16) is
1/16*(5+10+5+10)= 1/16* 30 = 15/8
and in total we get 15/8 + 15/8 = 30/8 = 15/4 or 3.75

if we do that for 100 shots, the expected value is 100*3.75 or 375 points

Answer 17

Wow amazingly explained thank you so much. This will help me so much!