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Geometric Probability

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Basic Geometric Probabilities

A rectangular dartboard that measures 12 inches by 24 inches has a 2-inch by 2-inch red square painted at its center. What is the probability that a dart that hits the dartboard will land in the red square?

Guidance

Sometimes we need to use our knowledge of geometry to determine the likelihood of an event occurring. We may use areas, volumes, angles, polygons or circles.

Example A

A game of pin-the-tale-on-the-donkey has a rectangular poster that is 2ft by 2ft. The area in which the tale should be pinned is shown as a circle with radius 1 inch. Assuming that the pinning of the tale is completely random and that it will be pinned on the poster (or the player gets another try), find the probability of pinning the tale in the circle?

Solution: This probability can be found by dividing the area of the circle target by the area of the poster. We must have the same units of measure for each area so we will convert the feet to inches.

\frac{1^2 \pi}{24^2} \thickapprox 0.005454 \ or \ \text{about} \ 0.5 \% \ \text{chance}.

Example B

In a game of chance, a pebble is dropped onto the board shown below. If the radius of each of blue circle is 1 cm, find the probability that the pebble will land in a blue circle.

Solution: The area of the square is 16 \ cm^2 . The area of each of the 16 circles is 1^2 \pi=\pi . The probability of the pebble landing in a circle is the sum of the areas of the circles divided by the area of the square.

P(\text{blue circle}) = \frac{16 \pi}{64} \thickapprox 0.785

Example C

What is the probability that a randomly thrown dart will land in a red area on the dart board shown? What is the probability that exactly two of three shots will land in the red? The radius of the inner circle is 1 unit and the radius of each annulus is 1 unit as well.

Solution: First we need to determine the probability of landing in the red. There are four rings of width 1 and the radius of the center circle is also 1 so the total radius is 5 units. The area of the whole target is thus 25 \pi square units. Now, we need to find the areas of the two red rings and the red circular center. The center circle area is \pi square units. The outside ring area can be found by subtracting the area inside from the entire circle’s area. The inside circle will have a radius of 4 units, the area of the outer ring is 25 \pi -16 \pi=9 \pi square units. This smaller red ring’s area can be found similarly. The circle with this red ring on the outside has a radius of 3 and the circle inside has a radius of 2 so, 9 \pi -4 \pi=5 \pi square units. Finally, we can add them together to get the total red area and divide by the area of the entire target. \frac{9 \pi+5 \pi+ \pi}{25 \pi}=\frac{15 \pi}{25 \pi}=\frac{3}{5} . So the probability of hitting the red area is \frac{3}{5} or 60%.

For the second part of the problem we will use a binomial probability. There are 3 trials, 2 successes and the probability of a success is 0.6: \dbinom{3}{2}(0.6)^2(0.4)=0.432

Intro Problem Revisit This probability can be found by dividing the area of the red square by the area of the dartboard. The area of the dartboard is 12\times24=288 . The area of the red square is 2\times2=4 . Therefore, the probability of the dart landing in the red square is

\frac{4}{288}\\\frac{1}{72}\\\approx 0.0139

Therefore, there is about a 1.39% chance the dart will hit the red square.

Guided Practice

1. Consider the picture below. If a “circle” is randomly chosen, what is the probability that it will be:

a. red

b. yellow

c. blue or green

d. not orange

2. If a dart is randomly thrown at the target below, find the probability of the dart hitting in each of the regions and show that the sum of these probabilities is 1. The diameter of the center circle is 4 cm and the diameter of the outer circle is 10 cm. What is the probability that in 5 shots, at least two will land in the 4 region?

Answers

1. a. \frac{29}{225}

b. \frac{69}{225}

c. \frac{84}{225}

d. \frac{182}{225}

2. P(1)&=\frac{2^2 \pi}{5^2 \pi}=\frac{4}{25} \\P(2)&=\frac{120}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{3} \times \frac{21 \pi}{25 \pi}=\frac{7}{25} \\P(3)&=\frac{90}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{4} \times \frac{21 \pi}{25 \pi}=\frac{21}{100} \\P(4)&=\frac{150}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{5}{12} \times \frac{21 \pi}{25 \pi}=\frac{35}{100}

Now add them up:

& P(1) + P(2) + P(3) + P(4) = \\&=\frac{4}{25}+\frac{7}{25}+\frac{21}{100}+\frac{35}{100} \\&=\frac{16}{100}+\frac{28}{100}+\frac{21}{100}+\frac{35}{100} \\&=\frac{100}{100}=1

The probability of landing in region 4 at least twice in five shots is equivalent to 1 - \left [ P(0) + P(1) \right ] .

Use binomial probability to determine these probabilities:

&1-\left [ \dbinom{5}{0} \left(\frac{35}{100}\right)^0 \left(\frac{65}{100}\right)^5+ \dbinom{5}{1} \left(\frac{35}{100}\right)^1 \left(\frac{65}{100}\right)4\right ] \\&=1-(0.116029+0.062477) \\& \thickapprox 0.821

Practice

Use the diagram below to find the probability that a randomly dropped object would land in a triangle of a particular color.

  1. yellow
  2. green
  3. plum
  4. not yellow
  5. not yellow or light blue

The dart board to the right has a red center circle (bull’s eye) with area \pi \ cm^2 . Each ring surrounding this bull’s eye has a width of 2 cm. Use this information to answer the following questions.

  1. Given a random throw of a dart, what is the probability that it will land in a white ring?
  2. What is the probability of a bull’s eye?
  3. What is the probability that in 10 throws, exactly 6 land in the black regions?
  4. What is the probability that in 10 throws, at least one will land in the bull’s eye?
  5. How many darts must be thrown to have a 95% chance of making a bull’s eye?

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