Independent Events (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Testing Independence with a Contingency Table

500 people were questioned and classified according to age and whether or not they regularly use social media. The results are shown in the table below.

Question: Is the regular use of social media independent of the respondent's age?

Solution:

To test for independence, we need to check whether $\text{Pr}(\text{Age} < 25 \cap \text{Yes}) = \text{Pr}(\text{Age} < 25) \times \text{Pr}(\text{Yes})$.

From the table:

Now calculate the product of the individual probabilities:

Since $\text{Pr}(\text{Age} < 25 \cap \text{Yes}) \neq \text{Pr}(\text{Age} < 25) \times \text{Pr}(\text{Yes})$:

Therefore, these events are not independent. This means that age does affect whether someone regularly uses social media.

Worked Example: Testing Multiple Pairs for Independence

An experiment consists of drawing a number at random from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$.

Let $A = \{1, 2, 3, 4\}$, $B = \{1, 3, 5, 7\}$ and $C = \{4, 6, 8\}$.

Question:

• a Are $A$ and $B$ independent?
• b Are $A$ and $C$ independent?
• c Are $B$ and $C$ independent?

Solution:

First, we calculate the individual probabilities:

a Testing independence of $A$ and $B$:

Find the intersection: $A \cap B = \{1, 3\}$

Calculate the product:

Since $\text{Pr}(A) \times \text{Pr}(B) = \text{Pr}(A \cap B)$, we conclude that $A$ and $B$ are independent.

b Testing independence of $A$ and $C$:

Find the intersection: $A \cap C = \{4\}$

Calculate the product:

Since $\text{Pr}(A) \times \text{Pr}(C) \neq \text{Pr}(A \cap C)$, we conclude that $A$ and $C$ are not independent.

c Testing independence of $B$ and $C$:

Find the intersection: $B \cap C = \varnothing$ (the empty set - they have no elements in common)

Calculate the product:

Since $\text{Pr}(B) \times \text{Pr}(C) \neq \text{Pr}(B \cap C)$, we conclude that $B$ and $C$ are not independent.

Worked Example: Distinguishing Physical from Mathematical Independence

Suppose we roll a die twice and define the following events:

• $A$ = the first roll shows a 4
• $B$ = the second roll shows a 4
• $C$ = the sum of the numbers showing is at least 10

Question:

• a Are $A$ and $B$ independent events?
• b What about $A$ and $C$?

Solution:

a Testing independence of $A$ and $B$:

Events $A$ and $B$ are physically independent (the result of one die roll doesn't affect the other), so they must also be mathematically independent. However, let's verify this.

Calculate the product:

The sample space for two dice rolls can be written as ordered pairs $(x, y)$ where $x$ is the first roll and $y$ is the second roll:

The total number of outcomes is $n(\varepsilon) = 36$.

The event $A \cap B$ corresponds to rolling a 4 on both dice: $(4, 4)$.

Therefore, $n(A \cap B) = 1$.

This confirms that $A$ and $B$ are independent.

b Testing independence of $A$ and $C$:

First, identify event $C$. The outcomes where the sum is at least 10 are:

Therefore, $n(C) = 6$.

Calculate the product:

The event $A \cap C$ represents outcomes where the first roll is a 4 AND the sum is at least 10. Looking at event $C$, the only outcome that satisfies this is $(4, 6)$.

Therefore, $n(A \cap C) = 1$.

This means that $A$ and $C$ are also independent events, even though they are not physically independent. The first roll being a 4 and the sum being at least 10 are related events, but mathematically they satisfy the independence condition.

Worked Example: Applying the Multiplication Rule

The probability that a family in a certain town owns a television set ($T$) is 0.75, and the probability that a family owns a station wagon ($S$) is 0.25. Assuming these events are independent, find the following probabilities:

a A family chosen at random owns both a television set and a station wagon.

b A family chosen at random owns at least one of these items.

Solution:

a Finding the probability of owning both items:

The event 'owns both a television set and a station wagon' is represented by $T \cap S$.

Since $T$ and $S$ are independent:

Therefore, there is an 18.75% probability that a randomly chosen family owns both items.

b Finding the probability of owning at least one item:

The event 'owns at least one of these items' is represented by $T \cup S$.

Using the addition rule:

Since $T$ and $S$ are independent, we can use $\text{Pr}(T \cap S) = \text{Pr}(T) \times \text{Pr}(S)$:

Therefore, there is an 81.25% probability that a randomly chosen family owns at least one of these items.