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**emmakatecumbo****Member**- Registered: 2017-09-22
- Posts: 35

Here is my homework I have been stuck for a couple days and need help. I listed below the teachers comments along with the diagram. Thank you <3

1. Robin knows that if he hits the white part of the target, he just slightly win, therefore not embarrassing John. What is the probability that Robin will hit the white part of the target (not the red and not the bull’s-eye)? Show your work.

3.14(2)2

= 12.56

3.14(6)2

= 113.04

3.14(9)2

= 254.34

2. After Robin shoots into the white area, John knows he cannot win, so he decides to set his sites on second place. To get second place, he needs to hit either the white circle or the bull’s-eye. What is the probability that he can do that? Show your work.

3.14(2)2 =

12.56 / /

3.14(6)2 =

113.04 / /

3.14(9)2 =

254.34

Problem B. The following is a diagram of a shuffleboard table

The table is 30 inches wide and 180 inches long. The width of the 2 and 3 rectangles is 12 inches. The sides of the 4 rectangle are 15 inches and the top and bottom are 12 inches. The 1 rectangle is twice as wide as the 2

3. What are the dimensions of the 1 rectangle?

4. What are the measurements of one of the 5 rectangles?

5. You are playing shuffleboard in P.E. class. To score points, your disc must land in a box on the other side of the table and you are awarded the number of points that is in that box. What is the probability of scoring 1 point? 1 out of 5

6. What is the probability of scoring 2 points? 1 out of 5

7. What is the probability of scoring 3 points?1 out of 5

8. What is the probability of scoring 4 point?1 out of 5

9. What is the probability of scoring 5 point?1 out of 5

10. What is the probability of scoring at all (any number of points)? 5 out of 5

Teacher Comments :

#1-2 you need to find the PROBABILITY... take the specific area asked for and divide by area of entire target!

#3-4 where are these? Use the info given to LABEL the diagram

$5-10 find the area of each rectangle and then use those too find probability!

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi emmakatecumbo

Hurrah! At last you have given me a diagram for the shuffleboard. Now we can make some progress.

You have used 'postimage' for the diagram. I have a problem with that. (1) The image shows on a page that is full of adverts. MIF has to keep its pages suitable for children to view. Some of the adverts I saw are not. I use imgur for my images. The posters to imgur sometimes post unsuitable material too, but the link I provide on my posts is to the image only; you would have to go to the imgur site to see other material. That makes it better for MIF. So I have copied your diagram to my imgur account and replaced your image link with mine. I'll do the same with your other post when I get time. But I'd prefer that you do this yourself as it takes time.

Here's how to get a picture (on its own) into a thread:

http://www.mathisfunforum.com/viewtopic … 17631&p=23 post number 1686

Now to the problem.

3. What are the dimensions of the 1 rectangle?

4. What are the measurements of one of the 5 rectangles?

Both these can be worked out with some simple arithmetic.

You're told that the '1' is twice the width of the '2' so that's easy to calculate.

If you then add up all the widths and subtract from 108, you'll have the width of the open (non scoring) space.

The '4' space is 15 top to bottom, so subtracting from 30 will give you the remainder for the two '5s'. So divide that remainder by 2 to get the missing dimension of the '5'.

Now sketch the whole board and by length x width write in the area of every section. The overall area is 30 x 108.

The probability of landing in any space is (area of that space) divided by (overall area)

Hope that helps.

later edit:

Robin knows that if he hits the white part of the target, he just slightly win, therefore not embarrassing John. What is the probability that Robin will hit the white part of the target (not the red and not the bull’s-eye)? Show your work.

3.14(2)2

= 12.56

3.14(6)2

= 113.04

3.14(9)2

= 254.34

There must have been a diagram for this which you haven't included. You could at least have described the target in words eg. : the target is a circular board with a circular centre, a red ring around this and a white outer. The radius values are ??? ??? and ???

This question (and the shuffle board question) rely on two unstated assumptions: "The player always hits somewhere on the board and it is equally likely which region the dart hits. This assumption is not true in real life. If I throw a dart at a target there is a high probability it will hit the wall and drop on the floor.

But, with the assumptions, what we need to know is the area of the regions involved. Larger areas are more likely to be hit than smaller areas.

So I'll assume you can work out:

(1) area of bull's eye (A1)

(2) red area (A2) = area of red circle minus A1

(3) white area (A3)= area of whole target minus red area.

Now the probability of landing a dart in a region is:

If you want further help on this it is essential] that you post back the radius of each circle and the calculations leading to the probabilities. I can then see if what you are doing is correct and either tell you the good news or tell you where you are going wrong.

Bob

*Last edited by bob bundy (2017-11-26 21:36:40)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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