Mastering Independent Events through Multiplication Rule and Sample Space

When there is more than one event in the same or a different sample space, the occurrence of one event either affects the occurrence of other events or it does not. This lesson will cover the independence of events.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Probability of an event
Experimental probability
Union
Intersection
Complement

Challenge

Selecting Two Students From a Group of Four Students

Emily is a student at North High and top of her class. Her mathematics teacher asks her to select two students and form a team of three students for the coming year's mathematics competition. When Emily sees the student list, she notices that the four students who come after her have almost identical grades in math.

The list of students with their math scores

She decides to select two students out the four possibilities, which includes one boy and three girls. Use a sample space to determine whether the following events are dependent or independent.

a Emily randomly selects a girl first, and she randomly selects a girl second.

b Emily randomly selects a boy first, and she randomly selects a girl second.

Explore

Investigating the Probability of Drawing a Marble

To comprehend the probabilities of events in different situations, the following exploration can be used. Suppose there are three marbles — one blue, one green, and one orange — in a bowl.

Consider drawing two marbles from the bowl randomly, one at a time.

Event A: Drawing a blue marble.
Event B: Drawing an orange marble.

In the following two cases, try to determine whether events A and B depend on each other or not.

a Calculate

P (A \cap B)

if the first marble is replaced before the second draw. Then analyze the product of

P (A)

and

P (B) .

Are the values the same or different?

b If the first marble is not replaced before the second draw, how does this change affect

P (A \cap B),

P (A),

and

P (B) ?

Compare the product of

P (A)

and

P (B)

with

P (A \cap B) .

Discussion

Independent and Dependent Events

Here the formal definitions of independent and dependent events will be presented.

Concept

Independent Events

Two events $A$ and $B$ are independent events if the occurrence of either of these events does not affect the occurrence of the other. It is also said that they are independent if and only if the probability that both events occur is equal to the product of the individual probabilities.

P (A \cap B) = P (A) \cdot P (B)

Why

For example, consider drawing two marbles from a bowl, one at a time.

Let

G,

B,

and

O

be the events of drawing green, blue, and orange marbles, respectively. The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. There is

1

green marble and

3

marbles in total.

P (G) = \frac{1}{3}

Suppose that the first marble is replaced before the second draw. Therefore, after the replacement there is

1

orange marble, and

3

marbles in total.

P (O) = \frac{1}{3}

Note that there are

9

possible outcomes for drawing two marbles one at a time. Only

1

of these options corresponds to an event of drawing a green marble and then an orange marble.

G G G B G O B B B G B O O O O G O B

Therefore, the combined probability of picking a green marble first and an orange marble second is

\frac{1}{9} .

Since the probability that both events occur is equal to the product of the individual probabilities, these events can be considered as independent events.

\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} ⇓ P (G) \cdot P (O) = P (G \cap O)

Concept

Dependent Events

Two events $A$ and $B$ are considered dependent events if the occurrence of either of these events affects the occurrence of the other. If the events are dependent, the probability that both events occur is equal to the product of the probability of the first event occurring and the probability of the second event occurring after the first event.

P (A \cap B) = P (A) \cdot P (B ∣ A)

Why

For example, consider drawing two marbles from a bowl, one at a time.

Let

G,

B,

and

O

1

green marble and

3

marbles in total.

P (G) = \frac{1}{3}

Suppose that after the green marble is picked, it is not replaced in the bowl.

This affects the probability of picking an orange marble on the second draw. Now there is still

1

orange marble, but instead of

3,

there are

2

marbles in total.

P (O ∣ G) = \frac{1}{2}

Using this information, the sample space of the described situation can be found.

G B G O B G B O O G O B

Out of

6,

there is only

1

outcome that corresponds to first drawing a green marble and then an orange marble. Therefore, the probability of picking a green and then an orange marble is

\frac{1}{6} .

\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} ⇓ P (G) \cdot P (O ∣ G) = P (G \cap O)

Because the occurrence of the first event affects the occurrence of the second, these events can be concluded to be dependent.

Example

Finding the Probability of Two Events Occurring at the Same Time Using the Multiplication Rule of Probability

The formulas that define independent and dependent events are also known as the Multiplication Rule of Probability. When knowing whether events are independent or dependent, this rule can be used to find the probability of both events occurring at the same time. Assume that Davontay flips a coin and rolls a dice at the same time.

Help Davontay find the probability of obtaining heads and an even number. Write the answer as a simplified fraction.

Hint

Begin by finding the probability of each event separately. Then use the Multiplication Rule of Probability.

Solution

There are two possible outcomes when flipping a coin, heads and tails. With this information, the probability of obtaining heads can be found by dividing favorable outcomes by the possible outcomes.

P (H) = \frac{1}{2}

Conversely, there are six possible outcomes when rolling a die,

1,

2,

3,

4,

5,

and

6 .

Only three of them are favorable outcomes —

2,

4,

and

6 .

P (E) = \frac{3}{6} \Leftrightarrow P (E) = \frac{1}{2}

Since flipping a coin and rolling a die do not affect each other's outcomes, these two events are independent. Knowing this, the probability that both events occur at the same time can be found by using the Multiplication Rule of Probability.

P (H \cap E) = P (H) \cdot P (E)

SubstituteII

$P (H) = \frac{1}{2}$ , $P (E) = \frac{1}{2}$

P (H \cap E) = \frac{1}{2} \cdot \frac{1}{2}

MultFrac

Multiply fractions

P (H \cap E) = \frac{1}{4}

Example

Rolling a Die and Deciding Whether the Given Events Are Independent

Davontay wants to practice the independence of events. To do so, he uses a balanced six-sided die.

He rolls the die and considers the following events.

A = {1, 2} B = {2, 3, 4, 5, 6}

Based on the definitions of independent and dependent events, he tries to find out whether these events are independent or not. Help Davontay during his practice!

Hint

Start by calculating $P (A),$ $P (B),$ and $P (A \cap B) .$ Use the definition of independent events.

Solution

Recall that two events are independent if and only if the probability that both events occur is equal to the product of the individual probabilities.

P (A \cap B) = P (A) \cdot P (B)

To calculate these probabilities, begin by identifying the sample space. Since Davontay rolls a six-sided die, there are $6$ elements in the sample space.

S = {1, 2, 3, 4, 5, 6}

Next, to find the probability that both events occur, identify the common elements in each event.

A = {1, 2} B = {2, 3, 4, 5, 6}

Notice that there is only one element in common. Therefore, the number of favorable outcomes is $1$ and the number of possible outcomes is $6$ for $P (A \cap B) .$

P (A \cap B) = \frac{1}{6}

Now, the individual probabilities can be calculated proceeding in the same way. Since there are

2

and

5

elements in

A

and

B,

respectively, these are the number of favorable outcomes for each of these events.

P (A) = \frac{2}{6} P (B) = \frac{5}{6}

Finally, check whether these probabilities satisfy the definition of independent events.

P (A \cap B) = P (A) \cdot P (B)

SubstituteValues

Substitute values

\frac{1}{6} = ? \frac{2}{6} \cdot \frac{5}{6}

MultFrac

Multiply fractions

\frac{1}{6} = ? \frac{1 0}{3 6}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{6} \neq = \frac{5}{1 8}

Since the probabilities do not satisfy the definition, event $A$ and event $B$ are not independent. This implies that $A$ and $B$ are dependent events.

Pop Quiz

Identifying Dependent and Independent Events

Davontay decides that one exercise is not enough practice. He needs more practice with independent events. Given the probabilities that both events $A$ and $B$ occur, event $A$ occurs, and event $B$ occurs, help Davontay to decide whether $A$ and $B$ are independent or dependent events.

The values of P(A and B), P(A), and P(B) are randomly generated

Example

Deciding Whether the Given Events About Euro $2020$ Predictions Are Independent

After Davontay is done practicing, he decides to watch the predictions of the semi-finals of Euro $2020 .$ His favorite teams are England and Italy. He really wants to see these teams in the finals. According to the sports announcer, England has a $56 %$ chance of winning against Denmark.

Davontay gets excited to hear the predictions for his other favorite team. However, the broadcast freezes for a couple of seconds and he misses the projection for Italy. He only hears that the chance of England playing against Italy is $20.72 % .$

Davontay is quite curious about Italy's chances of winning against Spain. Keeping in mind that England's win does not affect Italy's win, help Davontay satisfy his curiosity.

Hint

Use the definition of independent events and the fact that England's and Italy's wins do not affect each other.

Solution

Let

E

be the event of England winning against Denmark. Let

I

be the event of Italy winning against Spain. From here,

P (E \cap I)

and

P (E)

need to be identified. Note that

P (E \cap I)

is the probability that both England and Italy win against their opponents. Otherwise, they will not play against each other. For simplicity, convert the percentages to decimals.

P (E \cap I) P (E) = 0.2072 = 0.56

Since England's and Italy's wins do not affect each other,

E

and

I

are independent events. Therefore, the probability that both England and Italy win against their opponents is equal to the product of the probability that England wins and the probability that Italy wins.

P (E \cap I) = P (E) \cdot P (I)

Since

P (E \cap I)

and

P (E)

are known, by substituting these values into this equation the probability that Italy wins against Spain can be found.

P (E \cap I) = P (E) \cdot P (I)

SubstituteII

$P (E \cap I) = 0.2072$ , $P (E) = 0.56$

0.2072 = 0.56 \cdot P (I)

▼

Solve for

P (I)

DivEqn

$LHS / 0.56 = RHS / 0.56$

\frac{0 . 2 0 7 2}{0 . 5 6} = P (I)

RearrangeEqn

Rearrange equation

P (I) = \frac{0 . 2 0 7 2}{0 . 5 6}

UseCalc

Use a calculator

P (I) = 0.37

Italy has a

0.37,

37 %

chance of winning.

Example

Deciding Whether the Given Events When Picking a Card Are Independent

Davontay is excited to see his favorite teams in the Euro $2020$ finals. After the match, he goes to the library to study probability with his friend. Davontay wants to help his friend understand the independent and dependent events. To do this, he puts $8$ cards on a desk.

Davontay shuffles these cards and picks two cards, one at a time. Note that he replaces the first card before picking the second card.

Two cards are picked and returned to the deck

Based on this, he defines two different cases for his friend. Help his friend answer these questions.

a Let

A

be the event that the first selected card is an ace, and

S

be the event that the second selected card is a spade. Are the events

A

and

S

independent?

b Let

Q

be the event that the first selected card is a queen, and

H

be the event that the second selected card is a heart. Are the events

Q

and

H

independent?

Hint

a Calculate the number of favorable outcomes for each event by calculating the number of aces of spades, aces, and spades.

b Use the definition of independent events.

Solution

a By the definition of independent events, if the probability of the selected card being an ace of spades is equal to the product of the probability of the selected card being an ace and the probability of the selected card being a spade, then the events are independent.

P (A \cap S) = P (A) \cdot P (S)

Davontay's friend should begin by calculating

P (A \cap S),

P (A),

and

P (S) .

Notice that there are

8

cards and only one of them is the ace of spades. With this information,

P (A \cap S)

can be calculated by dividing the favorable outcomes by the possible outcomes.

P (A \cap S) = \frac{1}{8}

Using the same method, $P (A)$ and $P (S)$ can be also calculated.

P (A) = \frac{2}{8} P (S) = \frac{4}{8}

From here, by substituting these values into the equation it can be determined whether event $A$ and event $S$ are independent.

P (A \cap S) = P (A) \cdot P (S)

SubstituteValues

Substitute values

\frac{1}{8} = ? \frac{2}{8} \cdot \frac{4}{8}

MultFrac

Multiply fractions

\frac{1}{8} = ? \frac{8}{6 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 8}{b / 8}$

\frac{1}{8} = \frac{1}{8}

Therefore, the events are independent.

b Following the same procedure, Davontay's friend can decide whether events

Q

and

H

are independent. Notice that there are

2

queens,

2

hearts, and

1

queen of hearts. With this information,

P (Q \cap H),

P (Q),

and

P (H)

can be calculated as follows.

	Favorable Outcomes	Total Outcomes	Substitute
$P (Q \cap H)$	$1$	$8$	$\frac{1}{8}$
$P (Q)$	$2$	$8$	$\frac{2}{8}$
$P (H)$	$2$	$8$	$\frac{2}{8}$

Now that the probabilities have been calculated, Davontay's friend can find out whether the events are independent or dependent.

P (Q \cap H) = P (Q) \cdot P (H)

SubstituteValues

Substitute values

\frac{1}{8} = ? \frac{2}{8} \cdot \frac{2}{8}

MultFrac

Multiply fractions

\frac{1}{8} = ? \frac{4}{6 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

\frac{1}{8} \neq = \frac{1}{1 6}

Since the probability that both events will occur is not equal to the product of the individual probabilities of the events,

Q

and

S

are dependent events.

Pop Quiz

Calculating Probability of an Event Using Independency of Events

After playing with the cards, Davontay's friend also wonders if the definition of independent events can be used to find one of the individual probabilities given that the events are independent. Therefore, Davontay provides several exercises for his friend, knowing that the definition of independent events is biconditional.

Let $A$ and $B$ be independent events. Given the probability that both event $A$ and event $B$ occur and the probability that event $A$ occurs, find the probability that event $B$ occurs.

Values of P(A and B), P(A) are given, P(B) is asked to calculated

Closure

Deciding Whether the Events Are Independent or Dependent

This lesson has covered how to determine whether two events are independent or dependent. Using this knowledge, the challenge provided at the beginning of the lesson can be solved. Recall that Emily is selecting two students out of one boy and three girls who have the greatest math grades after Emily.

Use a sample space to determine whether the following events are dependent or independent.

a Emily randomly selects a girl first and she randomly selects a girl second.

b Emily randomly selects a boy first and she randomly selects a girl second.

Hint

a Begin by identifying the sample space for the events. Then use the definition of independent events.

b Identify the events whose probabilities should be found. What are the favorable and the total number of outcomes for each of these events?

Solution

a Given that two students will be selected out of one boy and three girls, the sample space of the events can be found. Let

B

represent the boy and

G_{1},

G_{2},

and

G_{3}

represent the girls.

Number of Girls	Outcome
$111222$	$G_{1} B G_{2} B G_{3} B G_{1} G_{2} G_{1} G_{3} G_{2} G_{3}$	$B G_{1} B G_{2} B G_{3} G_{2} G_{1} G_{3} G_{1} G_{3} G_{2}$

Therefore, there are $12$ outcomes in total. The events can be also identified as follows.

$A :$ Emily randomly selects a girl first.
$B :$ Emily randomly selects a girl second.

By the definition of independent events, if $P (A \cap B)$ is the product of $P (A)$ and $P (B),$ then $A$ and $B$ are said to be independent events.

P (A \cap B) = P (A) \cdot P (B)

Analyzing the sample space, it can be seen that there are

9

outcomes in which a girl is chosen first,

9

outcomes in which a girl is chosen second, and

6

outcomes in which two girls are chosen. Using this information, the values of

P (A),

P (B),

and

P (A \cap B)

can be found.

P (A) P (B) P (A \cap B) = \frac{9}{1 2} = \frac{9}{1 2} = \frac{6}{1 2}

Having found the probabilities, substitute these values into the equation.

P (A \cap B) = P (A) \cdot P (B)

SubstituteValues

Substitute values

\frac{6}{1 2} = ? \frac{9}{1 2} \cdot \frac{9}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 6}{b / 6}$

\frac{1}{2} = ? \frac{9}{1 2} \cdot \frac{9}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

\frac{1}{2} = ? \frac{3}{4} \cdot \frac{3}{4}

MultFrac

Multiply fractions

\frac{1}{2} \neq = \frac{9}{1 6}

Therefore, in this case events

A

and

B

are dependent.

b The second case be also examined in the same way. Begin by identifying the events.

$A :$ Emily randomly selects a boy first.
$B :$ Emily randomly selects a girl second.

Next, calculate the probabilities of these events.

P (A) P (B) P (A \cap B) = \frac{3}{1 2} = \frac{9}{1 2} = \frac{3}{1 2}

Now, substitute these probabilities into the equation.

P (A \cap B) = P (A) \cdot P (B)

SubstituteValues

Substitute values

\frac{3}{1 2} = ? \frac{3}{1 2} \cdot \frac{9}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

\frac{1}{4} = ? \frac{1}{4} \cdot \frac{3}{4}

MultFrac

Multiply fractions

\frac{1}{4} \neq = \frac{3}{1 6}

As it can be seen, the events in the second case are also dependent.

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Catch-Up and Review

Why

Why

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution