Two events are said to be independent if the occurrence (or non-occurrence) of one event does not affect the probability that the other event will occur.

Example.

Coins toss as an example of independent events

Suppose we toss a coin twice and we define the following events:

event $A$ - obtaining a heads for the first toss
event $B$ - obtaining a heads for the second toss

Because the outcome of the first coin toss does not affect the outcome of the next coin toss, these events are independent.

Example.

Relationship between outcome of football match and outcome of coin toss

Suppose we have the following events:

event A - the weather is rainy
event B - obtaining a heads for a single coin toss

Clearly, these events are independent because the weather has absolutely no impact on the probability of a tossed coin resulting in heads.

Theorem.

Multiplication rule of independent events

If $A$ and $B$ are independent events, then:

$$\mathbb{P}(A\cap{B})= \mathbb{P}(A)\cdot\mathbb{P}(B)$$

Proof. Rearranging the formula for conditional probability:

$$\begin{equation}\label{eq:KNmyJfG7wjdt6LKwHg2} \mathbb{P}(A\cap{B})=\mathbb{P}(A|B)\cdot\mathbb{P}(B) \end{equation}$$

If events $A$ and $B$ are independent, then this means that probability of $A$ is not affected by the occurrence of $B$, which means that $\mathbb{P}(A\vert{B}) = \mathbb{P}(A)$. Therefore \eqref{eq:KNmyJfG7wjdt6LKwHg2} becomes:

$$\begin{equation}\label{eq:nIOEIGHvKWmtcy3pnJp} \mathbb{P}(A\cap{B})=\mathbb{P}(A)\cdot\mathbb{P}(B) \end{equation}$$

■

Example.

Rolling two fair dices

Suppose we roll two fair dices. What is the probability that the outcome of both the dices is even?

Solution. Let events $A$ and $B$ represent the outcome of the first and second dice respectively. Events $A$ and $B$ are independent because the outcome of the first dice does not affect outcome of second dice. Therefore, the probability that the outcome of both dices is even is given by:

$$\begin{align*} \mathbb{P}(A\text{ is even }{\cap}\; B\text{ is even}) &=\mathbb{P}(A\text{ is even})\cdot\mathbb{P}(B\text{ is even})\\ &=\frac{3}{6}\cdot\frac{3}{6}\\ &=\frac{1}{4} \end{align*}$$

■

Difference between independent events and disjoint events

There is a common misconception that independent events and disjoint events are equivalent. Disjoint events are events that cannot occur simultaneously, and are illustrated in the diagram below:

For instance, suppose we toss a coin once and define the following two events:

event $A$ - outcome of toss is heads.
event $B$ - outcome of toss is tails.

Clearly, these events cannot happen at the same time - only one of the two events can occur and not both. Now, the question is, are events $A$ and $B$ independent? These two events are highly dependent because if we know the result of event $A$, then we will definitely know the result of event $B$.

Published by Isshin Inada

Edited by 0 others

Did you find this page useful?

thumb_up

thumb_down

Comment

Citation

Ask a question or leave a feedback...

thumb_up

thumb_down

chat_bubble_outline

settings

Enjoy our search

Hit / to insta-search docs and recipes!