# Connection of independent, depended, mutually exclusive, and mutually events [closed]

• Let say we made experiment that gave us N results.
• Let say m results are favorable for event A.
• Let say n results are favorable for event B.

Then we could picture it like that for independent, depended, mutually exclusive, and mutually events: events

According to the picture:

• independent and mutually exclusive events are the same thing
• dependent and mutually events are the same thing

Questions:

What the difference between:

1. independent and mutually exclusive events
2. depended and mutually events

Is it possible that event will be independent and mutually exclusive at the same time? If yes how we can use it?

Is it possible that event will be dependent and mutually at the same time? If yes how we can use it?

## closed as unclear what you're asking by 5xum, Did, Namaste, The Phenotype, JMPFeb 2 '18 at 5:31

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• Where did you get the picture? The phrase "mutually events" is gramatically incorrect... – 5xum Feb 1 '18 at 13:30
• @5xum, I draw it according to mutually events description. So that it just how I understand it. Could you please point me where I am wrong? – No Name QA Feb 1 '18 at 13:32
• Possibly helpful: math.stackexchange.com/questions/1855104/… – Ethan Bolker Feb 1 '18 at 13:32
• "mutually events" is non-sensical. Also, your image is incorrect. Can you write the definition? – 5xum Feb 1 '18 at 13:33
• @5xum, Event А is mutually from event B, if probability of А depends on if B happened or not – No Name QA Feb 1 '18 at 13:38

Two events $A,B$ are mutually exclusive iff: $$A\cap B=\varnothing\tag1$$

In words: it cannot happen that both events occur. At most one of them will occur, so the occurrence of $B$ excludes the occurrence of $A$ and vice versa. A consequence of this concerning probability is that: $$P(A\cap B)=P(\varnothing)=0$$

Two events $A,B$ are independent iff: $$P(A\cap B)=P(A)\times P(B)\tag2$$

A consequence of this is: $$P(A\mid B)=P(A)=P(A\mid B^{\complement})$$

In words: the probability of the occurrence of event $A$ is not depending on the occurrence (or non-occurence) of event $B$.

There is no mathematical concept "mutually events".

Two events $A$ and $B$ are mutually exclusive and independent at the same time iff :$$A\cap B=\varnothing\text{ and }0\in\{P(A),P(B)\}$$

In words: $A$ and $B$ have no outcome in common and at least one of them has probability $0$ to occur.

• Yes. That case is handled in the last paragraph in my answer. An example: if $A=\varnothing$ then for any event $B$ it is true that the events $A$ and $B$ are independent and are also mutually exclusive. – drhab Feb 2 '18 at 20:54
• If events can be independent and mutually exclusive at the same time why we need both? Could we just use for example only mutually exclusive? – No Name QA Feb 2 '18 at 22:53
• It seems that you are not well aware of the fact that "mutually exclusive" and "independent" are two concepts that are essentially different. In almost all cases they even exclude each other. The special case where two events $A,B$ are independent and mutually exclusive is a very very rare case. It can happen that an animal is green and is a carnivoor as well. Is that for you a reason to put aside one of the concepts "green" and "carnivoor"? – drhab Feb 3 '18 at 9:43
• Accroding to definitions for independ and mutually exclusive events, we could say that (for both types of events): P(A + B) = P(A) + P(B). If I'm right than why we need both defenitions? – No Name QA Feb 3 '18 at 19:30
• You are not right. If $A$ and $B$ are mutually exclusive then $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)$ because $A\cap B=\varnothing$ and consequently $P(A\cap B)=0$. But if $A$ and $B$ are are independent then $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)\times P(B)$ because $P(A\cap B)=P(A)\times P(B)$. It seems that your mind is telling you over and over again that the concepts "mutually exclusive" and "independent" are the same. Do not listen to that anymore. The concepts are distinct, as becomes as clear as crystal if you read carefully my answer. – drhab Feb 3 '18 at 20:32