# If the radius of the shaded center circle is 2 and the radius of the entire dartboard is 4, what is the probability of throwing a dart and hitting the white part of the board? Round your answer to the nearest whole number.

## 75% 67% 25% 50%

Apr 13, 2017

In a blind throw, and assuming the dart must hit the board, the probability of hitting the shaded centre is 25%.

#### Explanation:

Since area is proportional to ${r}^{2}$, when one doubles the radius of a circle, as in this problem, the area increase by ${2}^{2}$ or 4.

So, the shaded area is only one-fourth of the larger area, and the probability of hitting it would appear to be one fourth (25%) as well.

Apr 15, 2017

3/4=75%

#### Explanation:

It is not clear from the question, which the 'white' part of the board is. However, as the middle circle is 'shaded', I assume that the white part is the unshaded part.

It is possible to find the comparison between the areas of the shaded and white parts without doing the whole calculations, but working just in terms of $\pi$.

Area shaded circle: $A = \pi {r}^{2}$ which is $A = \pi {\left(2\right)}^{2} = 4 \pi$

Area of the entire board: $A = \pi {\left(2 r\right)}^{2}$ which is $A = 16 \pi$

The white area is given by $16 \pi - 4 \pi = 12 \pi$

So, with a random throw which does hit the board,:

$P \left(\text{white}\right) = \frac{12 \pi}{16 \pi} = \frac{3}{4}$

3/4 = 75%