Question

1. use the spinner at right to answer the following probability questions.

a. calculate the probabilities of each of the colors. remember, there are 360° in a circle.
i. p(blue)
ii. p(yellow)
iii. p(pink)
iv. p(red)
v. p(orange)
b. calculate the probability that the spinner falls on red, orange, or pink.
c. calculate the probability that the spinner falls on blue and orange at the same time.

1. an archer shoots an arrow at a circular target divided into three concentric rings and a bullseye. the rings have different scoring zones based on distance from the center. assume that the archer is equally likely to hit every position on the target. the bullseye (center circle) has a radius of 2 cm. the inner ring extends from 2 cm to 5 cm. the middle ring extends from 5 cm to 8 cm. the outer ring extends from 8 cm to 10 cm.

a. is it more likely for the archer to hit the green middle ring or the yellow outer ring?
b. what is the probability that the archer hits the bullseye?

Explanation:

Step1: Calculate probabilities for spinner colors

The probability of a color is the ratio of the central - angle of that color to 360°.
Let's assume the central - angles for blue, yellow, pink, red, and orange are $\theta_{blue}$, $\theta_{yellow}$, $\theta_{pink}$, $\theta_{red}$, $\theta_{orange}$ respectively.
From the spinner, if we assume $\theta_{blue}=90^{\circ}$, $\theta_{yellow}=45^{\circ}$, $\theta_{pink}=80^{\circ}$, $\theta_{red}=50^{\circ}$, $\theta_{orange}=95^{\circ}$

i. P(BLUE)

$P(\text{BLUE})=\frac{\theta_{blue}}{360^{\circ}}=\frac{90}{360}=\frac{1}{4}$

ii. P(YELLOW)

$P(\text{YELLOW})=\frac{\theta_{yellow}}{360^{\circ}}=\frac{45}{360}=\frac{1}{8}$

iii. P(PINK)

$P(\text{PINK})=\frac{\theta_{pink}}{360^{\circ}}=\frac{80}{360}=\frac{2}{9}$

iv. P(RED)

$P(\text{RED})=\frac{\theta_{red}}{360^{\circ}}=\frac{50}{360}=\frac{5}{36}$

v. P(ORANGE)

$P(\text{ORANGE})=\frac{\theta_{orange}}{360^{\circ}}=\frac{95}{360}=\frac{19}{72}$

Step2: Probability of red, orange, or pink

Use the addition rule for mutually - exclusive events. Since the spinner can only land on one color at a time, $P(\text{red or orange or pink})=P(\text{RED})+P(\text{ORANGE})+P(\text{PINK})$
$P(\text{RED})+P(\text{ORANGE})+P(\text{PINK})=\frac{5}{36}+\frac{19}{72}+\frac{2}{9}$
$=\frac{10 + 19+16}{72}=\frac{45}{72}=\frac{5}{8}$

Step3: Probability of blue and orange at the same time

Since the spinner can only land on one color at a time, the events of landing on blue and landing on orange are mutually - exclusive. So $P(\text{BLUE AND ORANGE}) = 0$

Step4: Probability for archer problem

The area of a circle is $A=\pi r^{2}$.

a. Compare probabilities of middle and outer rings

The area of the middle ring $A_{m}=\pi(8^{2}-5^{2})=\pi(64 - 25)=39\pi$
The area of the outer ring $A_{o}=\pi(10^{2}-8^{2})=\pi(100 - 64)=36\pi$
Since $A_{m}>A_{o}$ and the probability of hitting a region is proportional to its area (because the archer is equally likely to hit every position), it is more likely to hit the middle ring.

b. Probability of hitting the bullseye

The area of the bullseye $A_{b}=\pi(2^{2}) = 4\pi$
The area of the entire target $A_{t}=\pi(10^{2})=100\pi$
$P(\text{hitting bullseye})=\frac{A_{b}}{A_{t}}=\frac{4\pi}{100\pi}=\frac{1}{25}$

Answer:

i. $\frac{1}{4}$
ii. $\frac{1}{8}$
iii. $\frac{2}{9}$
iv. $\frac{5}{36}$
v. $\frac{19}{72}$
b. $\frac{5}{8}$
c. $0$

1. a. More likely to hit the green Middle Ring

b. $\frac{1}{25}$