Geometric Probability

A selection is to be made between points G and F as seen below

Find the probability that selection falls in

1. GH
2. HF
3. FJ
4. GH or HF

Solution

Firstly the sample space between G and F has to be calculated.

From G to F, the distance $GF=3-(-5)=3+5=8.$

Hence, the sample space is $GF=8.$

a. In order to find the probability that the selection falls in GH, we calculate first the distance GH,

$GH=-2-(-5)=3$,

Hence,

$P\left(GH\right)=\frac{GH}{GF}=\frac{3}{8}.$

b. In order to find the probability that the selection falls in HF, we calculate the distance HF,

$HF=3-(-2)=3+2=5$,

Hence,

$P\left(HF\right)=\frac{HF}{GF}=\frac{5}{8}.$

c. In order to find the probability that the selection falls in FJ, and before calculating the distance FJ, let us note that the region FJ is not within GF, so the fact that FJ occurs in GF is zero, hence

$P\left(FJ\right)=\frac{0}{8}=0$.

d. In order to find the probability that the selection falls into GH or HF, it is a union of two occurring events. Thus, we have,

$P(GH\cup HF)=P\left(GH\right)+P\left(HF\right)-P(GH\cap HF)$

but, $P\left(GH\right)=\frac{3}{8}$, calculated in part a, $P\left(HF\right)=\frac{5}{8},$ calculated in part b.

We have $P(GH\cap HF)=0$, thus $P(GHUHF)=\frac{3}{8}+\frac{5}{8}=\frac{8}{8}=1.$

The Stagecoach bus runs every 15 minutes at Frimley. It takes John 10 minutes to get to school from the bus stop. If John arrives at the bus station at half-past seven and is expected to get to school at quarter to eight.

What is the probability that he reaches his school on time?

Solution

To find out John's chances of reaching school at the appropriate time we need to find the best time the bus can arrive for John to meet up with school at 7:45. To calculate this we need to know the longest time John would wait at the bus stop before he starts running late. We have,

Time from Bus stop to School = 10 minutes.

John gets to the bus stop at 7:30 and has to be in school at 7:45. This means that John still has 15 minutes (i.e. 7:45 - 7:30) to meet up.

If the bus were to come at exactly 7:30 which is when John gets to the bus station it means John would arrive school at 7:40 which makes him 5 minutes early. So, John can only wait till 7:35 (i.e. 7:30 + 5 minutes) before he gets late to school.

John's favorable waiting time is thus 5 minutes, and the maximum total time he could wait for the bus is 15 minutes.

Therefore,

$\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{he}\mathrm{reaches}\mathrm{his}\mathrm{school}\mathrm{on}\mathrm{time}=\frac{\mathrm{favourable}\mathrm{waiting}\mathrm{time}}{\mathrm{total}\mathrm{available}\mathrm{waiting}\mathrm{time}}\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{he}\mathrm{reaches}\mathrm{his}\mathrm{school}\mathrm{on}\mathrm{time}=\frac{5}{15}\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{he}\mathrm{reaches}\mathrm{his}\mathrm{school}\mathrm{on}\mathrm{time}=\frac{1}{3}$$P(Johngettingtoschoolontime)=\frac{favourablewaitingduration}{Maximumwaitingduration}=\frac{5}{15}=\frac{1}{3}$

A circular path has been divided into 3 regions, arcs A, B and C. Find the probability a seedling growing a circular path is found in arc B if the arc length of A is 5cm, the arc length of C is 12cm and the radius of the circular path is 7cm. Take $\mathrm{\pi }=\frac{22}{7}$.

Solution

To determine the probability of the seedling growing in B we need our sample space. The sample space is the total distance around the circular path.

The total distance around the circular path is equal to the circumference of the path. But

Circumference of the circular path$=2\mathrm{\pi r}=2\times \frac{22}{7}\times 7$,

so the total distance around the circular path is 44 cm.

Now, we need to know the arc length of B having given the arc lengths of of A and C as 5cm and 12cm respectively.

But the total distance around the circular path is 44cm, thus

$ArclengthofB=totaldis\tan ce-(A+C)=44-(5+12)=44-17=27cm.$

So the probability that the seedling is found in B is

$P\left(B\right)=\frac{ArclengthofB}{Arclengthofthecircularpath}=\frac{27}{44}.$

$P\left(B\right)=\frac{27}{44}$

The figure below is a rectangular lawn with a length 15cm and width 30cm. A equilateral triangle sandy court has been created inside the lawn. If a golfer hits a golf into the lawn, what is the probability that the golf hits the sandy court?

Solution

Step 1.

We find the area of the rectangular lawn.

Step 2.

We find the area of the equilateral triangular sandy court.

$Areaoftriangle=\frac{1}{2}\times base\times height=\frac{1}{2}\times 6\times 8=24c{m}^2$

Step 3.

We find the probability that the golf hits the sandy court.

$P(golfhitsthesandycourt)=\frac{Areaoftriangularsandycourt}{Areaofrectangularlawn}=\frac{24}{450}=\frac{4}{75}.$

The figure below is a target. If its longest and shortest radii are 56cm and 7cm respectively. Find the probability of an arrow not hitting the bull's eye (the red spot). Take $\mathrm{\pi }=\frac{22}{7}.$

Solution

Step 1.

We Find the area of the whole target. Since it is a circle, the longest radius is used to find its area.

$Areaoftarger=\pi r^2\mathit{=}\frac{22}{7}\times {56}^2=9856c{m}^2$

Step 2.

We find the area of the bull's eye (red region). Knowing that the bull's eye (red region) is the smallest circle, that means it has the smallest radius. Thus;

$Areaofbull\text{'}seye=\frac{22}{7}\times 7^2=154c{m}^2.$

Step 3.

We find the probability of hitting the bull's eye. Let B be the event hitting the bull's eye.

$P(Hittingthebull\text{'}seye)=P\left(B\right)=\frac{Areaofbull\text{'}seye}{Areaoftarget}=\frac{154}{9856}=\frac{1}{64}$

Step 4.

We find the probability of not hitting the bull's eye. So B' is the event of not hitting the bull's eye, hence

$P(Nothittingthebull\text{'}seye)=1-P(hittingthebull\text{'}seye)$, so

$P(B\text{'})=1-P\left(B\right)=1-\frac{1}{64}=\frac{63}{64}.$