# Independent VS. Mutually Exclusive Events

*Probability* and *chance* surround us every day. If you have ever watched the news, the weather and market trends are given as probabilities. The chance of rain or snow is a great example of probability in everyday life.

Before going over independent and mutually exclusive events, let’s review some important probability vocabulary.

### Vocabulary

Probability: The likeliness that an event will occur. It is measured by the favorable outcomes over the total possible outcomes.

$$P\;\left(event\right)\;=\;\frac{Number\;of\;favorable\;outcome}{Total\;number\;of\;possible\;outcomes}$$

Event: An event is an outcome or a set of outcomes of an experiment.

Sample Space: The set of all possible outcomes of an experiment.

## Types of Events

*A* *mutually exclusive* *event* is a situation in which two events cannot occur at the same time.

** Independent events **are when one event is unaffected by another event.

**are when one event affects the outcome of another event.**

*Dependent events*** Simple**: An event with a single outcome

**: Two independent events occurring. Found by multiplying the probabilities of the events.**

*Compound*Let’s take a look at flipping a coin. When you flip a coin there are two possible outcomes. One outcome is the coin lands on heads, the second outcome is that the coin lands on tails.

The *experiment* is flipping the coin. The *sample space* for this experiment is heads and tails.

When you flip a coin once you cannot get both heads and tails. This means that the events of getting heads or getting tails are ** mutually exclusive**.

This is also *a simple event* since there is one event that is taking place.

Notice that with the diagram, there is no overlap between the two events.

Now let’s say you are going to roll a standard six sided die and flip a coin.

Since there are two events this is now a compound event. If you land on four with the die, that does not affect what the coin will land on. These two events are ** independent**.

Keeping with the numbered die, two events that would not be mutually exclusive would be rolling an odd number and rolling a 5.

Rolling a 3 would satisfy one of the events, rolling an odd. However, if a 5 is rolled, it satisfies both events and happens at the same time.

Notice the overlap with these events. There are numbers that are odd that are not five, but a five is always odd.

## Did you know?

- Mutually exclusive events are always dependent.
- The probability of two people in a room of 23 people having the same birthday is 50%
- A probability of 0 means that the event is impossible
- A probability of 1 means that the event is a certainty.

## Probability Notation

What is the probability of rolling a 4 with a standard six sided die? Let’s start by looking at the sample space, the different numbers that the die can land on.

Sample Space: 1, 2, 3, 4, 5, 6

The favorable outcome for this experiment is 4.

$$\operatorname{𝑃}\;\left(event\right)\;=\;\frac{Number\;of\;favorable\;outcome}{Total\;number\;of\;possible\;outcomes}$$

$$\operatorname{𝑃}\;(rolling\;\operatorname{𝑎}\;4)\;=\;\frac16$$

### Example 1:

*What is the probability of flipping a tail and rolling an even number?*

$$\operatorname{𝑃}\;(tails\;and\;even)\;=\;?$$

Let’s start with the probability of flipping a **tail**.

$$P\;(tails)\;=\;\frac1{\;2}$$

Next, what is the probability of rolling an **even** number?

$$P\;\left(even\right)\;=\;\frac3{\;6}\;=\;\frac{\;1\;}2$$

To find the probability of these independent events, the probabilities are multiplied.

$$P\;\left(tails\;and\;even\right)\;=\;\frac1{\;2}\;\times\;\frac1{\;2}\;=\;\frac1{\;4}$$

### Example 2:

*What is the probability of flipping both a tail and heads with one coin?*

Since only one coin is being used, you can only get one result, this means that the events are mutually exclusive.

$$P\;\left(heads\;and\;tails\right)\;=\;0$$

### Example 3:

*What is the probability of choosing an ace or a queen from a standard deck of cards?*

$$P\;\left(Ace\right)\;=\;\frac4{52\;}\;=\;\frac1{13}$$

$$P\;\left(Queen\right)\;=\;\frac4{52}\;=\;\frac1{13}$$

$$P\;\left(Ace\;OR\;Queen\right)\;=\;\frac1{13}\;+\;\frac1{13}\;=\;\frac2{13}$$

You can also look at the number of favorable outcomes, 8, over the total number of cards, 52. This will also reduce to become $$\frac2{13}$$ .

### Example 4:

Given a standard deck of cards.

Let H = choosing a heart

Let E = choosing an even number

Let F = choosing a five

Let A = choosing an Ace

*Which of the following are mutually exclusive events?*

Start with H and compare it to the other events.

- Can a card be both a heart and an even? Yes
- Can a card be both a heart and a five? Yes
- Can a card be both a heart and an Ace? Yes

Next look at E.

- Can a card be both an even and a five? No- mutually exclusive events
- Can a card be both an even and an Ace? No – mutually exclusive events

The last pairing is F and A

- Can a card be both a five and an ace? No – mutually exclusive events