# Nakafa Framework: LLM

URL: /en/subject/high-school/10/mathematics/probability/two-events-not-mutually-exclusive
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/10/mathematics/probability/two-events-not-mutually-exclusive/en.mdx

Output docs content for large language models.

---

export const metadata = {
  title: "Non-Mutually Exclusive Events A and B",
  description: "Calculate overlapping events using P(A or B) = P(A) + P(B) - P(A and B). Avoid double counting with intersection examples and step-by-step solutions.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "04/21/2025",
  subject: "Probability",
};

## What Does Non-Mutually Exclusive Mean?

We learned about [mutually exclusive events](/subject/high-school/10/mathematics/probability/two-events-mutually-exclusive) that can't happen together (like turning left and right at the same time). Now, let's talk about **Non-Mutually Exclusive Events**. These are two (or more) events that **CAN happen at the same time** in a single experiment.

This means it's possible to get an outcome that belongs to event A and also belongs to event B.

**Simple Examples:**

1.  **Drawing a Card:** You draw one card from a standard deck.

    - Event A: Getting a **Heart** (<InlineMath math="\heartsuit" />).
    - Event B: Getting a **King**.
      Can events A and B happen together? Absolutely! There's a card that is both a Heart and a King: the **King of Hearts** (<InlineMath math="K\heartsuit" />). Since they can happen together, events A and B are **non-mutually exclusive**.

2.  **Rolling a Die (once):**
    - Event A: Getting an **even** number (<InlineMath math="\{2, 4, 6\}" />).
    - Event B: Getting a number **greater than 3** (<InlineMath math="\{4, 5, 6\}" />).
      Can these happen together? Yes! The numbers <InlineMath math="4" /> and <InlineMath math="6" /> are both even and greater than 3. So, events A and B are **non-mutually exclusive**.

## The Intersection is Important!

In non-mutually exclusive events, there's a part that belongs to both events simultaneously. This part is called the **intersection**.

Because there is an intersection, the probability of event A **AND** B happening together is **greater than zero**.

<BlockMath math="P(A \text{ and } B) > 0" />

Or using the intersection symbol:

<BlockMath math="P(A \cap B) > 0" />

This is very different from mutually exclusive events, where <InlineMath math="P(A \cap B) = 0" />.

## Calculating P(A OR B) for Non-Mutually Exclusive Events

Since there's a chance that events A and B can happen together, we can't just add <InlineMath math="P(A) + P(B)" /> to find <InlineMath math="P(A \text{ or } B)" />.

Why not? Because if we simply add them, the intersection part (<InlineMath math="A \cap B" />) gets **counted twice**, once in <InlineMath math="P(A)" /> and again in <InlineMath math="P(B)" />.

To get the correct calculation, we must **subtract** the probability of the intersection that was double-counted. This gives us the **General Addition Rule** for probability:

<BlockMath math="P(A \text{ or } B) = P(A) + P(B) - P(A \cap B)" />

Or using the union and intersection symbols:

<BlockMath math="P(A \cup B) = P(A) + P(B) - P(A \cap B)" />

This formula works generally, for both mutually exclusive and non-mutually exclusive events. (If they are mutually exclusive, <InlineMath math="P(A \cap B)" /> is zero, so the formula simplifies back to <InlineMath math="P(A \cup B) = P(A) + P(B)" />).

## Calculation Example

Let's use the card example:

- Event A: Getting a Heart (<InlineMath math="\heartsuit" />). There are 13 Hearts in 52 cards. <InlineMath math="P(A) = 13/52" />.
- Event B: Getting a King. There are 4 Kings in 52 cards. <InlineMath math="P(B) = 4/52" />.
- Event A **and** B: Getting the King of Hearts (<InlineMath math="K\heartsuit" />). There is only 1 King of Hearts. <InlineMath math="P(A \cap B) = 1/52" />.

So, the probability of getting a Heart OR a King is:

<div className="flex flex-col gap-4">
  <BlockMath math="P(A \cup B) = P(A) + P(B) - P(A \cap B)" />
  <BlockMath math="= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}" />
  <BlockMath math="= \frac{16}{52} = \frac{4}{13}" />
</div>

See? We subtract <InlineMath math="1/52" /> so the King of Hearts isn't counted twice.