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8. Probability
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United States of AmericaTennessee
Algebra 2

8.01 Independent and dependent events

Lesson
Practice
Lesson

Concept summary

In probability, an event is a set of outcomes of an experiment to which a probability is assigned. Two events in the same experiment can be classified as either independent or dependent events.

Independent Events

Two events are independent if the occurence of one event does not affect the likelihood of the occurence of the other event.

Example:

The outcomes of flipping two coins

Dependent Events

Two events are dependent if the occurence of one event affects the likelihood of the occurence of the other event.

Example:

The outcomes of drawing two cards in a standard deck without replacement

Probability of Independent Events:

If two events, A and B, are independent, then the probability of both events occuring is the product of the probability of A and the probability of B:

\displaystyle P\left(A \cap B \right)=P\left(A\right) \cdot P\left(B\right)
\bm{P(A)}
Probability of the first event
\bm{P(B)}
Probability of the second event

Worked examples

Example 1

State whether the following events are independent or dependent:

a

A coin is tossed and a fair six-sided die is rolled.

Approach

To determine whether two events are independent or dependent, we should determine if the outcome of one event influences the outcome of the other event.

Solution

The outcome of tossing a coin, either heads or tails, does not affect the outcome of rolling a six-sided die. Therefore, the events are independent.

b

A card dealer randomly chose a card from a standard deck and hid it in his pocket. The deck is then shuffled and a new card is chosen.

Approach

To determine whether two events are independent or dependent, we should determine if the outcome of one event influences the outcome of the other event.

Solution

Since the dealer hid one of the 52 cards, there are fewer cards to choose from when the second card is chosen. Therefore, the events are dependent.

Example 2

For each of the following scenarios, use probability to determine if the events are independent or dependent:

a

Stella spins a color spinner with three equally sized sections labeled G, Y and R twice. The first spin lands on Y and the second spin lands on G.

Approach

To find the sample space of spinning the spinner twice, a tree diagram should be constructed as shown:

A tree diagram with two levels. In the first level, there are three choices: G, Y and R. Each choice from the first level is connected by lines to three choices: G, Y and R.

From the tree diagram, the sample space has 9 equally likely outcomes: \{(G, G), (G, Y), (G, R), (Y, G), (Y, Y), (Y, R), (R, G), (R, Y), (R, R) \}

To test the independence of the two events we must determine if P(Y \cap G) = P(Y) \cdot P(G) is true.

Solution

\displaystyle P(Y \cap G)\displaystyle =\displaystyle \dfrac{1}{9}Only one event (Y, G) out of 9 satisfies this event
\displaystyle P(Y) \cdot P(G)\displaystyle =\displaystyle \dfrac{1}{3} \cdot \dfrac{1}{3}P(Y)=P(G)= \dfrac{1}{3} since there are 3 colors
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{9}Evaluate
\displaystyle \therefore P(Y \cap G)\displaystyle =\displaystyle P(Y) \cdot P(G)Since both equal \dfrac{1}{9}

Therefore the two events are independent.

b

30 dancers audition for a part. The judges decide that 16 dancers have the right height and 20 dancers are good dancers. The events in this scenario are Right height, R, and Good dancer, G.

Approach

\\ n(R)+n(G) = 16+20 =36 \gt 30, so 6 dancers must be in R \cap G.

We can display the number of people in each category in a Venn diagram as shown:

A Venn diagram with an outer rectangle and two intersecting circles. The circle on the left is labelled as Right height and has a 10 inside it and a 6 inside the part that intersects with the circle on the right. The circle on the right is labelled as Good dancer and has a 14 inside it in addition to the 6 inside the intersection with the other circle.

To test the independence of the two events we must determine if P(R \cap G) = P(R) \cdot P(G) is true.

Solution

From the Venn diagram we see that: P(R)=\dfrac{16}{30}, \, P(G)=\dfrac{20}{30} and P(R \cap G)=\dfrac{6}{30}.

Determining if the two events are independent:

\displaystyle P(R \cap G)\displaystyle =\displaystyle \dfrac{1}{5}Simplified
\displaystyle P(R) \cdot P(G)\displaystyle =\displaystyle \dfrac{16}{30} \cdot \dfrac{20}{30}Substitute the probabilities
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{16}{45}Evaluate
\displaystyle \therefore P(R \cap G)\displaystyle \neq\displaystyle P(R) \cdot P(G)

Since P(R \cap G) is not equal to the product of P(R) and P(G), the two events are dependent.

Outcomes

A2.S.CP.A.1

Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. Categorize events as independent or dependent.*

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP6

Attend to precision.

A2.MP8

Look for and express regularity in repeated reasoning.

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