Ellie M.

asked • 02/24/15

Carnival games wooden ring

To win the carnival game Ring Ding, you must toss a wooden ring onto a grid of rectangles so that it lands without touching any of the grid lines. The ring has a 3-inch diameter, the rectangles are twice as long as they are wide, and the game has been designed so that you have a 28% chance of winning. What are the dimensions of each rectangle?

1 Expert Answer

By:

Mark M. answered • 02/24/15

Tutor
5.0 (278)

Mathematics Teacher - NCLB Highly Qualified

See tutors like this
See tutors like this
probability = successes / total
 
In this case the success is all of the ring (its area) within the rectangle (its area).
 
0.28 = Aring / Arectangle
0.28 = 1.52π / Arectangle
Arectangle = 1.52π / 0.28
Arectangle = 25.24
 
Arectangle = 2w2
25.24 = 2w2
12.6 = w2
3.5 ≈ w
7 ≈ l
Upvote • 0 Downvote
More
Report

Kevin K.

I have to disagree with your answer because I don't believe that you are taking into consideration the fact that the ring is circular so thus the basic area ratio won't work.  
 
I looked at it this way:  
1.  You have to assume that the ring will land on the board and as a result, I chose to focus just on one rectangle of the board since they are all the same size.  
 
2.  For the ring to land entirely inside the rectangle, the center of the ring needs to be 1.5 inches inside the border (radius is 1.5 inches), so the "winning rectangle" will have dimensions of (x - 2(1.5)) and (2x - 2(1.5)) or x-3 by 2x-3.
 
3.  Now, we will set up and solve our geometric probability equation.  The winning rectangle's area of (x-3)(2x-3) divided by the total rectangle's area of x(2x) = 28%
 
(2x-3)(x-3)/(2x2) = .28
2x2 - 9x + 9 = .56x2
1.44x2 - 9x + 9 = 0
 
Now you can plug that in the quadratic formula and get the solutions x = 1.25 or 5.  However, it cannot by 1.25 because that is too small of a dimension for the the circle with diameter of 3 to fit in, the answers has to be 5.
 
As a result, the dimensions of the rectangles are 5in x 10in.   
 
Report

06/22/16

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.