Question

A sporting goods store is offering its customers a chance to win a gold club. Customers who make a purchase are given one chance to win by throwing a golf ball at a board in the shape of a regular hexagon with 6-inch sides. In the center of the hexagon is a circular hole with a 6-inch diameter. Before throwing the ball, contestants must choose whether they will throw the ball through the hole or hit the wooden portion of the target. Which target provides the highest probability of winning? The circular hole or the solid part of the hexagon? Explain please

Answer 1

Answer:

The solid part of the hexagon

Step-by-step explanation:

To determine which provides the best chances, we have to look at the area of each of the two zones.

First, let's calculate the circle area, which follows the following formula:

$A = \frac{\pi D^{2} }{4} = \frac{\pi * 6^{2} }{4} = \frac{36\pi }{4} = 28.27$

So, the circle has an area of 28.27 square inches.

Now, let's calculate the hexagon area, which is obtained with this formula (s = side of the hexagon):

$A = \frac{3}{2} * \sqrt{3} * s^{2} = \frac{3}{2} * \sqrt{3} * 6^{2} = 54 \sqrt{3} = 93.53$

The TOTAL area of the hexagon is 93.53 inches.

To get the area of the solid part of the hexagon, we have to subtract the area of the circle: 93.53 - 28.27 = 65.26 square inches.

So the solid part of the hexagon is more than twice the size of the hole, making it much more probable to hit it.  On top of that, don't forget that if the center of the ball thrown is just inside the hole, the rest of the ball will most likely hit the hexagon frame, which increases even more the probability to hit the solid part, more than just the area ratio.