﻿ Geometric Probability Worksheet | Problems & Solutions
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# Geometric Probability Worksheet

Geometric Probability Worksheet
• Page 1
1.
Bus #R5711 runs every 35 minutes to Sam's college. If he arrives at the bus stop at a random time, then what is the probability that he will have to wait for 15 minutes or more? Assume that the bus stops for 5 minutes at the stop.
 a. $\frac{3}{4}$ b. $\frac{5}{7}$ c. $\frac{3}{8}$ d. $\frac{5}{8}$

#### Solution:

Divide the interval of 35 + 5 = 40 minutes into intervals of 5 minutes.

GH represents the time duration for which the bus waits at the stop.

If Sam arrives at the instant O, he has to wait for 40 minutes.

If Sam arrives at the instant E, he has to wait for 15 minutes.

The occurrence of the event of Sam waiting for 15 minutes or more is represented by OE.

Probability of Sam waiting for 15 minutes or more = OE / OH = 25 / 40 = 5 / 8
[From figure.]

Correct answer : (4)
2.
A dartboard is shown in the figure. Find the probability of hitting the shaded region. Given $a$ = 4 units, $b$ = 16 units.[π = 3]

 a. $\frac{1}{194}$ b. $\frac{1}{192}$ c. $\frac{1}{193}$ d. $\frac{1}{191}$

#### Solution:

Probability of hitting the shaded region = Area of the shaded regionArea of the board

Area of the shaded region = 42 - 2 π(2)22 = 4 units
[Area of square - Area of 2 semi circles.]

Area of the board = π × 162 = 768 units

Probability of dart hitting the target = 4 / 768 = 1 / 192

Correct answer : (2)
3.
Which of the following geometric probabilities is impossible?
 a. 0.45 b. 1.2 c. 0 d. 1

#### Solution:

As probability always lies between 0 and 1, geometric probability can never be greater than 1.

Correct answer : (2)
4.
What is the probability of getting 75 points when the spinner is spun?

 a. $\frac{1}{3}$ b. 1 c. $\frac{5}{18}$ d. $\frac{2}{9}$

#### Solution:

P(getting 75 points) = Angle subtended by arc CD total angle

P (getting 75 points) = 100o360o
[Substitute.]

P(getting 75 points) = 5 / 18
[Simplify.]

Correct answer : (3)
5.
A dart thrown at a rectangular board ABCD is equally likely to land on any point. Find the probability of the dart hitting the inner rectangle PQRS. [Given $a$ = 21 cm, $b$ = 16 cm, $c$ = 18 cm, $d$ = 10 cm.]

 a. $\frac{29}{15}$ b. $\frac{15}{28}$ c. $\frac{4}{7}$ d. $\frac{28}{15}$

#### Solution:

Area of the rectangle ABCD = AB × BC = 21 × 16 = 336 cm 2
[Area of rectangle = length × breadth.]

Area of the rectangle PQRS = PQ × QR = 18 × 10 = 180 cm 2
[Area of the rectangle = Length × breadth.]

P (Dart hitting the rectangle PQRS) = Area of the rectangle PQRS / Area of the rectangle ABCD.

The probability of the dart hitting the inner rectangle PQRS = 180 / 336 = 15 / 28.

Correct answer : (2)
6.
A dart thrown at a line segment $\stackrel{‾}{\mathrm{AC}}$ can land at any point on it. What is the probability that it lands on $\stackrel{‾}{\mathrm{BC}}$? [Given $a$ = 19, $b$ = 9.]

 a. $\frac{5}{14}$ b. $\frac{9}{19}$ c. $\frac{9}{28}$ d. $\frac{9}{29}$

#### Solution:

P (Dart landing on BC) = Length ofBCLength ofAC.

P (Dart landing on BC) = 99+19 = 9 / 28
[Substitute and simplify.]

Correct answer : (3)
7.
A dart is thrown at a board in the form of a circle. If the dart hits the board, then what is the probability that it will land in the shaded area? [Given $x$° = 66o.]

 a. $\frac{49}{60}$ b. $\frac{37}{45}$ c. $\frac{149}{180}$ d. $\frac{73}{90}$

#### Solution:

P (Dart landing in the shaded area) = area of the shaded portionarea of the circle.

Required probability = area of the sector AOCDAarea of the circle.

Required probability = Angle subtended by arc CDAtotal angle.
[Area is proportional to the central angle.]

Required probability = 3600-6603600 = 294 / 360 = 49 / 60
[Substitute and simplify.]

Correct answer : (1)
8.
Equilateral triangle PQR, of side 36 in. is inside a square ABCD of side 48 in.. A dart is thrown at the square board. If the dart lands on the board and that it is equally likely to land on any point, then the probability of hitting the triangle PQR is

 a. $\frac{9\sqrt{3}}{64}$ b. $\frac{18\sqrt{3}}{64}$ c. $\frac{11\sqrt{3}}{64}$ d. $\frac{9}{64}$

#### Solution:

Area of the square ABCD = (48)2 = 2304 in.2
[Area of a square = (side)2.]

Area of the ΔPQR = 34 × (36)2 = 3243 in.2
[Area of an equilateral triangle = 34 × (side)2.]

P(Dart hitting ΔPQR) = area of ΔPQR area of square ABCD

Required probability = 32432304 = 9364
[Substitute and simplify.]

Correct answer : (1)
9.
A dart is thrown at a square ABCD. What is the probability that it lands in the shaded region?

 a. $\frac{2}{5}$ b. $\frac{1}{5}$ c. $\frac{1}{4}$ d. $\frac{1}{3}$

#### Solution:

By symmetry, area of ΔAOB = 1 / 4(Area of ABCD)

P (Dart landing in the shaded portion AOB) = Area of the shaded portion AOBArea of the square ABCD

Required probability = 1 / 4

Correct answer : (3)
10.
A dart is thrown at ΔABC. What is the probability that it lands on shaded portion MNK?

 a. $\frac{1}{8}$ b. $\frac{1}{32}$ c. $\frac{1}{16}$ d. $\frac{1}{24}$

#### Solution:

Area of ΔPQR = 1 / 4(Area of ΔABC)
[By symmetry.]

Area of ΔMNK = 1 / 4(Area of ΔPQR)
[By symmetry.]

Area of ΔMNK = 1 / 16(Area of ΔABC)
[Step1 and Step2.]

Area of ΔMNK Area of ΔABC = 1 / 16
[Step3.]

P (Dart landing on shaded portion MNK) = area of ΔMNKarea of ΔABC = 1 / 16
[Step 4.]

Correct answer : (3)

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