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URL: https://nakafa.com/en/subjects/mathematics/probability/two-events-not-mutually-exclusive
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Calculate overlapping events using P(A or B) = P(A) + P(B) - P(A and B). Avoid double counting with intersection examples and worked solutions.

---

## What Does Non-Mutually Exclusive Mean?

We learned about [mutually exclusive events](/en/subjects/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 $$B$$.

Visible text: This means it's possible to get an outcome that belongs to event and also belongs to event .

**Simple Examples:**

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

    - Event $$A$$: Getting a **Heart** ($$\heartsuit$$).
    - 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** ($$K\heartsuit$$). Since they can happen together, events $$A$$ and $$B$$ are **non-mutually exclusive**.

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

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

 - Event : Getting a **Heart** ().
 - Event : Getting a **King**.
 Can events and happen together? Absolutely! There's a card that is both a Heart and a King: the **King of Hearts** (). Since they can happen together, events and are **non-mutually exclusive**.

2. **Rolling a Die (once):**
 - Event : Getting an **even** number ().
 - Event : Getting a number **greater than** ().
 Can these happen together? Yes! The numbers and are both even and greater than . So, events and 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**.

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

```math
P(A \text{ and} B) > 0
```

Or using the intersection symbol:

```math
P(A \cap B) > 0
```

This is very different from mutually exclusive events, where $$P(A \cap B) = 0$$.

Visible text: This is very different from mutually exclusive events, where .

## Calculating Combined Probability for Non-Mutually Exclusive Events

Since there's a chance that events $$A$$ and $$B$$ can happen together, we can't just add $$P(A) + P(B)$$ to find $$P(A \text{ or} B)$$.

Visible text: Since there's a chance that events and can happen together, we can't just add to find .

Why not? Because if we simply add them, the intersection part ($$A \cap B$$) gets **counted twice**, once in $$P(A)$$ and again in $$P(B)$$.

Visible text: Why not? Because if we simply add them, the intersection part () gets **counted twice**, once in and again in .

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:

```math
P(A \text{ or} B) = P(A) + P(B) - P(A \cap B)
```

Or using the union and intersection symbols:

```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, $$P(A \cap B)$$ is zero, so the formula simplifies back to $$P(A \cup B) = P(A) + P(B)$$).

Visible text: This formula works generally, for both mutually exclusive and non-mutually exclusive events. (If they are mutually exclusive, is zero, so the formula simplifies back to ).

## Calculation Example

Let's use the card example:

- Event $$A$$: Getting a Heart ($$\heartsuit$$). There are $$13$$ Hearts in $$52 \text{ cards}$$. $$P(A) = 13/52$$.
- Event $$B$$: Getting a King. There are $$4$$ Kings in $$52 \text{ cards}$$. $$P(B) = 4/52$$.
- Event $$A$$ **and** B: Getting the King of Hearts ($$K\heartsuit$$). There is only $$1$$ King of Hearts. $$P(A \cap B) = 1/52$$.

Visible text: - Event : Getting a Heart (). There are Hearts in . .
- Event : Getting a King. There are Kings in . .
- Event **and** B: Getting the King of Hearts (). There is only King of Hearts. .

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

Component: MathContainer
Children:

```math
P(A \cup B) = P(A) + P(B) - P(A \cap B)
```

```math
= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}
```

```math
= \frac{16}{52} = \frac{4}{13}
```

See? We subtract $$1/52$$ so the King of Hearts isn't counted twice.

Visible text: See? We subtract so the King of Hearts isn't counted twice.