PRACTICE PROBLEMS ON PROBABILITY

Problem 1 :

An urn contains 12 balls : 

3 Red, 5 Yellow and 4 Black

A- In this part we draw successfully and without replacement 3 balls from this urn.

Calculate the probability of the following :

D : the three balls are red.

E : the 3 balls have the same color.

F1 : the 3 balls red, Yellow, Black in this order .

F2 : the 3 balls are have different color.

G1: If the 3 balls are Red Red Black in this order.

G2 : If the 3 balls are 2 red and 1 Black.

H : If the 3 balls don't have the same color.

Solution :

Total number of balls = 3 Red + 5 Yellow + 4 Black

D : the three balls are red.

Three balls are red, 

3C₃ = 1

Doing simplification, we get

= 1/220

E : the 3 balls have the same color.

F1 : the 3 balls Red, Yellow, Black in this order.

P(RYB) = P(R) x P(Y) x P(B)

= (3/12) x (5/11) x (4/10)

=  1/22

F2 : the 3 balls are have different color.

G1: If the 3 balls are Red Red Black in this order.

P(RRB) = P(R) x P(R) x P(B)

= (3/12) x (2/11) x (4/10)

= 1/55

G2 : If the 3 balls are 2 red and 1 Black.

= (3C₂ x 4C₁)/12C3

= (3x4)/220

= 3/44

H : If the 3 balls don't have the same color.

= 1 - Probability of same color

= 1 - (3/44)

= 41/44

Problem 2 :

When two six-sided dice are rolled, there are 36 possible outcomes, as shown. Find the probability of each event.

a. The sum is not 6

b. The sum is less than or equal to 9.

Solution :

Total number of outcomes n(S) = 36

Let A be the event of getting 6 in the die.

a) Number of times getting the sum as 6 

n(A) = 5

P(A) = n(A)/n(S)

= 5/36

P(A') = 1 - (5/36)

= (36 - 5)/36

= 31/36

b) Let B be the event of getting the number greater than 9

Then B' be the event of getting the output less than or equal to 9.

The possible value of B are = 10, 11, 12

n(B) = 6

P(B) = n(B)/n(S)

= 6/36

= 1/6

P(B') = 1 - 1/6

= 5/6

Problem 3 :

You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you more likely to get 10 points or 0 points?

Solution :

Area of circle = πr²

Side length of the square = 18 inches

Area of square = 18²

= 324

Area of yellow portion = Area of large circle - area of green circle

Radius of yellow circle = 9 inches

Radius of green circle = 6 inches

Radius of pink circle = 3 inches

Area of pink portion = π(3²)

= 9π

= 9(3.14)

= 28.26

Probability of getting 10 points = 28.26/324

= 0.087222....

Approximately 0.0873

Probability of getting 0 points = Area of square - area of yellow portion

= [324 -  π(9²)]/324

= [324 - 81(3.14)]/324

= [324 - 254.34]/324

= 69.66/324

= 0.215

0.215 > 0.0873

So, you are more likely to get 0 points.

Problem 4 :

Each section of the spinner shown has the same area. The spinner was spun 20 times. The table shows the results. For which color is the experimental probability of stopping on the color the same as the theoretical probability?

Solution :

Total = 5 + 9 + 3 + 3

= 20

From the table :

Probability of getting yellow = 3/20

Probability of getting blue = 3/20

Probability of getting red = 5/20 ==> 1/4

Probability of getting green = 9/20

For experimental probability :

Probability of getting yellow = 1/4

Probability of getting blue = 1/4

Probability of getting red = 1/4

Probability of getting green = 1/4

So, the probability of getting red using experimental and theoretical are the same.

Problem 5 :

In the United States, a survey of 2184 adults ages 18 and over found that 1328 of them have at least one pet. The types of pets these adults have are shown in the fi gure. What is the probability that a pet-owning adult chosen at random has a dog?

Solution :

The number of trials is the number of pet-owning adults, 1328. A success is a pet-owning adult who has a dog. From the graph, there are 916 adults who said that they have a dog.

P(pet-owning adult has a dog) = 916/1328

= 229/332

= 0.689

Converting into percent, we get

= 0.689 x 100%

= 68.9

≈ 0.690

The probability that a pet-owning adult chosen at random has a dog is about 69%.