Mathematics (0580) | Probability of combined events

🌟 Probability of Combined Events: Your Ultimate Study Guide 🌟

Welcome to the exciting world of Combined Probability! So far, you’ve mastered calculating the chance of a single event happening—like rolling a 6 on a dice. But what if you roll the dice *twice*? Or draw two cards? That’s where combined events come in!

This chapter teaches you the essential rules and tools (like Venn and Tree Diagrams) to handle situations where two or more things happen together. Understanding this is key to solving many real-world problems, from gambling fairness to quality control in manufacturing.

1. Quick Review: Fundamentals of Probability

Before combining events, let’s quickly remind ourselves of the basics. Don't skip this—it’s the foundation!

1.1 Calculating Single Event Probability

The probability of an event \(A\) is always between 0 and 1 (or 0% and 100%).

$$P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$

1.2 The Complement Rule

The probability of an event happening plus the probability of it not happening must equal 1.

If \(A'\) (read as 'A prime' or 'A dash') is the event that $A$ does not happen, then:

$$P(A') = 1 - P(A)$$

Example: If the probability of rain \(P(R)\) is 0.4, the probability of no rain \(P(R')\) is \(1 - 0.4 = 0.6\).

2. Combining Independent Events (The "AND" Rule)

This is where things get combined! We first look at events that don't affect each other.

2.1 What are Independent Events?

Two events, A and B, are independent if the occurrence of A has absolutely no effect on the probability of B occurring.

• Analogy: If you flip a coin and roll a dice, the coin result doesn't change the chances of rolling a 6. They are separate actions.
• Common Context: Often involves replacement (e.g., drawing a marble and putting it back) or separate random devices (dice, coins).

2.2 The Multiplication Rule for Independent Events

To find the probability that Event A AND Event B both occur, you multiply their individual probabilities.

$$P(A \text{ and } B) = P(A) \times P(B)$$

We sometimes write $P(A \text{ and } B)$ as \(P(A \cap B)\) in the context of sets (Extended students, remember this notation!).

Did you know? Because you are multiplying two numbers less than 1, the combined probability will always be smaller than the individual probabilities. This makes sense—it’s harder to achieve two specific things than just one!

Step-by-Step Example (Independent Events)

A spinner has \(P(\text{Red}) = 0.2\) and a dice has \(P(\text{Odd}) = 0.5\). What is the probability of spinning Red AND rolling an Odd number?

1. Identify the probabilities: \(P(R) = 0.2\), \(P(O) = 0.5\).
2. Apply the multiplication rule (since they are independent):
   $$P(R \text{ and } O) = P(R) \times P(O)$$ $$P(R \text{ and } O) = 0.2 \times 0.5$$
3. Calculate the result: \(P(R \text{ and } O) = 0.1\).

3. Combining Mutually Exclusive Events (The "OR" Rule)

When you want to know the probability of one thing OR another happening.

3.1 What are Mutually Exclusive Events?

Two events are mutually exclusive if they cannot happen at the same time. There is no overlap.

• Analogy: You can roll a 4 on a standard dice, or you can roll a 5. You cannot roll both a 4 and a 5 in a single roll.
• Non-example: Drawing an Ace OR drawing a Red card is not mutually exclusive, because the Ace of Hearts is both red and an Ace (there is overlap).

3.2 The Addition Rule for Mutually Exclusive Events

To find the probability that Event A OR Event B occurs (when they are mutually exclusive), you add their individual probabilities.

$$P(A \text{ or } B) = P(A) + P(B)$$

Step-by-Step Example (Mutually Exclusive)

A bag contains 5 blue, 3 green, and 2 red marbles (total 10). What is the probability of picking a Blue OR a Red marble?

1. Identify individual probabilities: \(P(B) = \frac{5}{10} = 0.5\), \(P(R) = \frac{2}{10} = 0.2\).
2. Determine if they are mutually exclusive: Yes, a marble cannot be both blue and red at the same time.
3. Apply the addition rule:
   $$P(B \text{ or } R) = P(B) + P(R)$$ $$P(B \text{ or } R) = 0.5 + 0.2$$
4. Calculate the result: \(P(B \text{ or } R) = 0.7\).

4. Visualizing Combined Events

Sometimes formulas are tricky to apply directly. We have three excellent tools to map out all possible outcomes.

4.1 Sample Space Diagrams

A sample space diagram (or table) lists every possible combination of outcomes. This works best when there are only two combined events with a relatively small number of outcomes (like rolling two dice).

Example: Rolling two standard six-sided dice.

The sample space has 6 rows and 6 columns, totaling $6 \times 6 = 36$ possible outcomes.

• You can easily calculate \(P(\text{Sum is 7})\) by counting the combinations that equal 7 (there are 6 of them: (1,6), (2,5), etc.). \(P(\text{Sum is 7}) = \frac{6}{36}\).
• You can calculate \(P(\text{Two different numbers})\) by excluding the 6 pairs where the numbers are the same (1,1), (2,2)... \(P = \frac{36 - 6}{36} = \frac{30}{36}\).

4.2 Tree Diagrams

Tree diagrams are indispensable, especially for sequences of events. Outcomes are listed at the end of the branches, and probabilities are written along the branches.

How to Use Tree Diagrams:

1. Draw the first set of branches for the first event (e.g., Coin 1: Heads or Tails).
2. From the end of each first branch, draw the next set of branches for the second event (e.g., Coin 2: Heads or Tails).
3. Write the probability on each branch.
4. To find the probability of a path (AND): Multiply the probabilities along the path.
5. To find the probability of multiple successful outcomes (OR): Add the probabilities of the successful paths.

Example: Flipping a coin twice (Independent Events).
Path 1 (HH): \(P(H) \times P(H) = 0.5 \times 0.5 = 0.25\)
Path 2 (HT): \(P(H) \times P(T) = 0.5 \times 0.5 = 0.25\)
Probability of getting at least one Head? Add the probabilities of the paths containing H (HH, HT, TH): \(0.25 + 0.25 + 0.25 = 0.75\).

4.3 Venn Diagrams (Limited to Two Sets)

Venn diagrams help visualize overlapping or distinct events, particularly in problems involving characteristics or categories (like students who study Maths OR Physics).

The syllabus limits these to two sets, A and B. Remember the notation:

• Universal Set (U): Everything in the box.
• Intersection (\(A \cap B\)): The overlap area. This represents "A AND B".
• Union (\(A \cup B\)): All of A plus all of B. This represents "A OR B (or both)".
• Complement (\(A'\)): Everything outside set A.
• \(n(A)\): The number of elements in set A.

If you are given the number of students who like Maths, the number who like Physics, and the number who like BOTH, use a Venn diagram to solve it. Always start filling the diagram from the middle intersection.

5. Extended Content: Dependent Events (Without Replacement)

For students tackling Extended content, you must be comfortable with events where the probabilities change after the first event occurs. These are known as dependent events.

5.1 What are Dependent Events?

Two events, A and B, are dependent if the outcome of Event A changes the probability of Event B.

• Common Context: Drawing items without replacement (e.g., picking two socks from a drawer without putting the first one back).

5.2 Using Tree Diagrams for Dependent Events

Tree diagrams are crucial here, but you must be careful to update the probabilities on the second set of branches.

Step-by-Step Example (Dependent Events)

A bag contains 5 Red and 3 Blue marbles (Total 8). Two marbles are drawn without replacement. Find the probability of drawing two Blue marbles.

Event 1: Drawing the first marble

• \(P(\text{1st is Blue}) = \frac{3}{8}\)
• \(P(\text{1st is Red}) = \frac{5}{8}\)

Event 2: Drawing the second marble (AFTER the first one is removed)

Assume the first was Blue (B): Now there are 7 marbles left, and only 2 of them are Blue.

• \(P(\text{2nd is Blue | 1st was B}) = \frac{2}{7}\)

Calculate combined probability:

We want \(P(\text{B AND B})\):

$$P(B_1 \text{ and } B_2) = P(B_1) \times P(B_2 \text{ given } B_1)$$ $$P(B_1 \text{ and } B_2) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$$

Simplify: \(P(\text{B and B}) = \frac{3}{28}\)