Question

1. the target at right is made of three concentric circles, with their dimensions shown in the diagram in inches.

a. what is the probability that a randomly chosen point in the target is in the shaded region?
b. what is the probability that a randomly selected point in the target is in the two innermost circles?
c. what is the probability that a randomly selected point in the target is not in the smallest circle?

1. the diagram to the right depicts a triangle inscribed in a square. what is the probability that a randomly selected point in the square is not in the shaded regions?

challenge: try these if you want to push yourself a bit. it’s ok if you don’t do them. they aren’t required.

1. find the probability that a randomly selected point lies in the shaded region.
2. there are 800 students in our school. find the probability that at least two people share the same birthday.

Explanation:

Step1: Recall probability formula for geometric - probability

The probability \(P\) that a randomly - chosen point lies in a particular region is given by \(P=\frac{\text{Area of the favorable region}}{\text{Area of the total region}}\).

Step2: Solve part (a) of question 5

The radii of the three concentric circles are \(r_1 = 1\), \(r_2=2\), and \(r_3 = 3\) (assuming outer - most radius is \(3\) as it is \(1 + 2\)). The area of a circle is \(A=\pi r^{2}\). The area of the total target (outer - most circle) is \(A_{total}=\pi r_3^{2}=\pi\times3^{2}=9\pi\). The area of the non - shaded part (innermost circle) is \(A_{1}=\pi r_1^{2}=\pi\times1^{2}=\pi\). The area of the shaded region is \(A_{shaded}=A_{total}-A_{1}=9\pi-\pi = 8\pi\). So the probability \(P_a=\frac{A_{shaded}}{A_{total}}=\frac{8\pi}{9\pi}=\frac{8}{9}\).

Step3: Solve part (b) of question 5

The area of the two innermost circles: The area of the first circle with radius \(r_1 = 1\) is \(A_1=\pi r_1^{2}=\pi\times1^{2}=\pi\), and the area of the second circle with radius \(r_2 = 2\) is \(A_2=\pi r_2^{2}=\pi\times2^{2}=4\pi\). The combined area of the two innermost circles is \(A = A_1+A_2=\pi + 4\pi=5\pi\). The area of the total target is \(A_{total}=\pi\times3^{2}=9\pi\). So the probability \(P_b=\frac{5\pi}{9\pi}=\frac{5}{9}\).

Step4: Solve part (c) of question 5

The area of the smallest circle is \(A_{small}=\pi r_1^{2}=\pi\times1^{2}=\pi\), and the area of the total target is \(A_{total}=\pi\times3^{2}=9\pi\). The probability that a point is in the smallest circle is \(P_{in - small}=\frac{\pi}{9\pi}=\frac{1}{9}\). The probability that a point is not in the smallest circle is \(P_c=1 - P_{in - small}=1-\frac{1}{9}=\frac{8}{9}\).

Step5: Solve question 6

The area of the square is \(A_{square}=s^{2}\), where \(s = 5\), so \(A_{square}=25\) square inches. The area of the non - shaded triangle: The base and height of the non - shaded triangle are equal to the side of the square, so \(A_{triangle}=\frac{1}{2}\times s\times s=\frac{1}{2}\times5\times5=\frac{25}{2}\) square inches. The probability \(P_6=\frac{A_{triangle}}{A_{square}}=\frac{\frac{25}{2}}{25}=\frac{1}{2}\).

Step6: Solve challenge 1

The area of the circle with radius \(r = 7\) is \(A_{circle}=\pi r^{2}=\pi\times7^{2}=49\pi\). The area of a regular hexagon with side length \(a\) (in a circle of radius \(r\), \(a = r\)) is \(A_{hexagon}=\frac{3\sqrt{3}}{2}a^{2}\). Since \(a = 7\), \(A_{hexagon}=\frac{3\sqrt{3}}{2}\times49\). The probability that a point lies in the shaded region (area of the circle minus area of the hexagon) is \(P_{challenge1}=\frac{\pi\times49-\frac{3\sqrt{3}}{2}\times49}{49\pi}=\frac{\pi-\frac{3\sqrt{3}}{2}}{\pi}=1-\frac{3\sqrt{3}}{2\pi}\).

Step7: Solve challenge 2

The probability that no two people share the same birthday:
For \(n = 800\) people, the number of ways to assign non - repeating birthdays out of 365 days is \(P(365,n)=\frac{365!}{(365 - n)!}\). The total number of ways to assign birthdays to \(n\) people is \(365^{n}\).
The probability that no two people share the same birthday is \(p=\frac{365\times364\times\cdots\times(365 - 799)}{365^{800}}\).
The probability that at least two people share the same birthday is \(P_{challenge2}=1 - p\). Since \(n=800>365\), by the pigeonhole principle, the probability \(P_{challenge2}=1\).

Answer:

a. \(\frac{8}{9}\)
b. \(\frac{5}{9}\)
c. \(\frac{8}{9}\)

1. \(\frac{1}{2}\)

Challenge 1. \(1-\frac{3\sqrt{3}}{2\pi}\)
Challenge 2. \(1\)