Exercise 40 Page 673

a To find the probability that the coin will land entirely in the circle, we will need the area of the circle and the area of the region. Let's look at that first.

The whole target is the square with an area of $8 \cdot 8 = 64 in^{2} .$ The favorable area, is not simple the area of the circle. The coin has to lie entirely with in the circle. Therefore, its center needs to land within a smaller inner circle.

Since the coin's radius is

\frac{1 5}{3 2},

the center of our coin has to be more than that distance from the circles edge. That means the center must lie within the circle of radius

1 - \frac{1 5}{3 2} = \frac{1 7}{3 2} .

Let's find the area of that circle.

A = π r^{2}

Substitute

$r = \frac{1 7}{3 2}$

A = π (\frac{1 7}{3 2})^{2}

CalcPow

Calculate power

A = π (\frac{2 8 9}{1 0 2 4})

Multiply

A \approx 0.8866

We can now find the probability that the coin will land completely within that circle by look at the ratio of the inner circle's area to area of the square.

\frac{A _{Inner Circle}}{A _{Square}} = \frac{0 . 8 8 6 6}{6 4}

Therefore, the probability that a quarter lands entirely in the circle in one random toss is about

1.4 % .

b From Part A, we learned that the probability of a quarter landing entirely in the circle is about

1 % .

This percentage means that if the coins randomly fall within the square, $1$ in every $100$ would land entirely within the circle. Therefore, we must toss $100$ coins to expect a prize.

Subchapter links

Lesson Check p.671

Practice and Problem-Solving Exercises p.671-673

Standardized Test Prep p.674

Mixed Review p.674