Independent Events Revision Notes for HSC SSCE Mathematics Advanced

Independent Events

What are independent events?

Independent events are events where the occurrence of one event has no effect on the probability of the other event occurring. In simpler terms, knowing the outcome of one event tells us nothing useful about the outcome of the other event.

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For example, when you roll a die and flip a coin, these are independent events. The number that appears on the die does not affect whether the coin lands on heads or tails. Similarly, the coin result doesn't influence the die outcome.

Mathematical definition of independence

Two events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other. This can be expressed mathematically in two equivalent ways:

$P(A|B) = P(A)$

This means the probability of $A$ given that $B$ has occurred equals the probability of $A$ alone. In other words, knowing that $B$ has happened doesn't change the probability of $A$ .

Similarly:

$P(B|A) = P(B)$

This means the probability of $B$ given that $A$ has occurred equals the probability of $B$ alone.

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Both conditions are mathematically equivalent ways of expressing independence. If one is true, the other must also be true. You can use either condition to test whether two events are independent.

Testing for independence

To verify whether two events are independent, we can use the conditional probability formula and check if it satisfies the independence condition.

Starting with the conditional probability formula:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

If events $A$ and $B$ are independent, then $P(A|B) = P(A)$ , so:

$P(A) = \frac{P(A \cap B)}{P(B)}$

Multiplying both sides by $P(B)$ :

$P(A \cap B) = P(A) \times P(B)$

This is the multiplication rule for independent events. It shows that if two events are independent, the probability of both occurring equals the product of their individual probabilities.

Similarly, we can show that $P(B|A) = P(B)$ also leads to the same result: $P(A \cap B) = P(A) \times P(B)$ .

chatImportant

The multiplication rule $P(A \cap B) = P(A) \times P(B)$ is both a test for independence and a calculation tool. Both conditions $P(A|B) = P(A)$ and $P(B|A) = P(B)$ lead to this same result, demonstrating they are equivalent ways of expressing independence.

Therefore, both conditions are equivalent ways of expressing independence.

The multiplication rule for independent events

For independent events, the probability of both events occurring together (their intersection) is found by multiplying their individual probabilities:

$P(A \cap B) = P(A) \times P(B)$

Where $P(A \cap B)$ represents the probability of both $A$ and $B$ occurring.

This rule is particularly useful for solving practical problems involving repeated trials, such as multiple coin flips, rolling dice multiple times, or selecting items independently.

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Critical Rule: This multiplication rule only works when events are independent. If events are dependent (one affects the other), you cannot simply multiply their probabilities. Always verify independence before applying this rule.

Worked example 1: Die and coin flip

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Worked Example: Die and Coin Flip

A die is rolled, and a coin is flipped. Let $A$ be rolling an even number, and $B$ be flipping heads.

Part a: Find $P(A \cap B)$

First, we identify that rolling a die and flipping a coin are independent events, so we can use the multiplication rule:

$P(A \cap B) = P(A) \times P(B)$

For event $A$ (rolling an even number):

$A = \{2, 4, 6\}$ from the sample space $\{H, T\}$ for the coin

$|A| = 3 \times 2 = 6$

$P(A) = \frac{6}{12} = \frac{1}{2}$

For event $B$ (flipping heads):

$B = \{1, 2, 3, 4, 5, 6\} \times \{H\}$

$|B| = 6 \times 1 = 6$

$P(B) = \frac{6}{12} = \frac{1}{2}$

Now apply the multiplication rule:

$P(A \cap B) = P(A) \times P(B)$

$= \frac{1}{2} \times \frac{1}{2}$

$= \frac{1}{4}$

We can verify this using the basic probability formula. The sample space has $|S| = 6 \times 2 = 12$ outcomes. The intersection $A \cap B = \{2, 4, 6\} \times \{H\}$ has $|A \cap B| = 3 \times 1 = 3$ outcomes.

$P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{3}{12} = \frac{1}{4}$

Both methods give the same answer, confirming our calculation.

Part b: Show that $A$ and $B$ are independent

To prove independence, we need to show that $P(A|B) = P(A)$ .

Using the conditional probability formula:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

$= \frac{\frac{1}{4}}{\frac{1}{2}}$

$= \frac{1}{4} \times 2$

$= \frac{1}{2}$

$= P(A)$

Since $P(A|B) = P(A)$ , we have proven that events $A$ and $B$ are independent.

Worked example 2: Defective items

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Worked Example: Defective Items

A machine produces items, with 2% being defective. Two items are selected independently.

Part a: Find the probability both are defective

Let $A$ and $B$ represent the first and second items being defective respectively.

We know that $P(A) = P(B) = 0.02$ (2% = 0.02).

Since the items are selected independently, we use the multiplication rule:

$P(A \cap B) = P(A) \times P(B)$

$= 0.02 \times 0.02$

$= 0.0004$

The probability that both items are defective is 0.0004 or 0.04%.

Part b: Find the probability at least one is defective

To find the probability of at least one defective item, we can use the complement rule. The complement of "at least one defective" is "neither defective."

First, find the probability that neither item is defective.

Let $\bar{A}$ represent the first item not being defective, and $\bar{B}$ represent the second item not being defective.

$P(\bar{A}) = 1 - P(A) = 1 - 0.02 = 0.98$

$P(\bar{B}) = 1 - P(B) = 1 - 0.02 = 0.98$

Since the selections are independent:

$P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B})$

$= (1 - P(A)) \times (1 - P(B))$

$= (1 - 0.02) \times (1 - 0.02)$

$= 0.98 \times 0.98$

$= 0.9604$

Now apply the complement rule to find the probability of at least one defective:

$P(\text{at least one}) = 1 - P(\text{neither})$

$= 1 - 0.9604$

$= 0.0396$

The probability that at least one item is defective is 0.0396 or approximately 3.96%.

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Key Points to Remember:

Independent events are events where one outcome does not affect the probability of the other occurring. Knowing one event's outcome tells you nothing about the other.
Test for independence using either $P(A|B) = P(A)$ or $P(B|A) = P(B)$ . If either condition holds, the events are independent.
Multiplication rule for independent events: $P(A \cap B) = P(A) \times P(B)$ . This only works when events are independent.
The multiplication rule is particularly useful for repeated trials and scenarios where events occur separately without influencing each other.
When finding probabilities of "at least one" occurring, often the complement rule combined with independence is the most efficient approach: find the probability of "none" occurring, then subtract from 1.

Independent Events (HSC SSCE Mathematics Advanced): Revision Notes