Two events that are independent do not have to be mutually exclusive, or disjoint.
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@anim1-2110 жыл бұрын
This is short and easy to understand! Thanks!!!
@saberjan49464 жыл бұрын
Great teacher , I found it really helpful.... keep it up
@haldur8612 жыл бұрын
Thx man, first picture/slide was really helpful!
@INSANEcuber8 жыл бұрын
A life saver you are.
@roltgen15 жыл бұрын
Thank you; great example
@maysterchief11 жыл бұрын
Sorry it wasn't helpful. The main thing to remember is that Mutually exclusive events are events that CANNOT happen at the same time.
@racinerobinson2 жыл бұрын
the explanation was great but the example was bit hard to relate to since I don't have common knowledge of a deck of cards.
@Truckito200711 жыл бұрын
Hey, for whatever it's worth, i thought this was very good. You can't please everyone :) keep up the good work, man. I'll be referring to your vids for help. Thanks.
@Pratyush1s9 жыл бұрын
THANKS, FOR SHORT AND CLEAR CONCEPT. In last Q- Are "face card" and "king" Independent or Mutually Exclusive ? A- Neither Independent nor Mutually Exclusive. P(Face Card/King)= 1 ; P(King)=12/52 Therefore, P(Face Card/King) not equal to P(King)-------- Hence not Independent. P(Face Card and King)= 4/52 i.e not equal to 0 ---------------- Hence not Mutually Exclusive.
@Rusnurmat8 жыл бұрын
Hi guys,if someone knows I have only one question, the JOINT and NON-DISJOINT events are the same thing or not? please help me with this. and if this possible, explain if NO then what is this...
@Mephistahpheles7 жыл бұрын
"Two events that are independent -do not have to- _cannot_ be mutually exclusive"
@denizacar90253 жыл бұрын
Not, quite, Sir. This is a false statement. Two events can be independent and mutually exclusive if one of the events is 0 (of course the 'given that' event can't be 0 as you can't divide to 0, but the other event could be 0). It is important to clarify that two events can be also neither independent nor mutually exclusive.
@quantstyle64482 жыл бұрын
They are two different concepts but why can't they can both exist together? Getting heads or tails on a coin flip is mutually exclusive. Two coin flips are independent. Both situations can exist together, if we're flipping the coin twice.
@brandonryan15883 жыл бұрын
After Im done with stats Im going to eat then smoke- 0:48
@damienfis60527 ай бұрын
Think i'm more confused then i was before watching this.
@PedroBarba19707 жыл бұрын
Math aside, can you clarify why a "red card" and a "spade" are dependent events?
@abnormal4327af536 жыл бұрын
Pedro Barba when the probability of any event changes due to another event, then the event is dependent event. When you know you have picked up a spade, you know right away it's not red, because spades always are always black . Thus by picking up a spade you change the probability of the card being a red to 0. As the event of picking up a spade changes the probability of the card being a red, being red isdependent to being spade.
@Black182heart3 жыл бұрын
@@abnormal4327af53 isn't that the same with everything? If I get head, I know for sure I didn't get tails.
@harisasif26393 жыл бұрын
@@Black182heart Two events A and B are dependent if the probability of event B changes due to occurrence of event A. In your case, tossing a coin once is just a singular event. However, if the coin was tossed twice, then if you get heads on your first toss coin it does not mean that you will get tails on your second coin toss. In other words, the two events are independent of one another.
@sebakoli67815 жыл бұрын
really helpful. thanks
@amodabsoul Жыл бұрын
God blessed you teacher 😍
@schang_lh4 жыл бұрын
So basically, if the two given variable are independent, then they're not disjoint, and vice versa?
@MrRyanli9 жыл бұрын
Love it !!!! :)
@spookybrojo71863 жыл бұрын
A spade and red card aren’t mutually exclusive because you can have a spade that’s red
@paranghrahimi9 жыл бұрын
p(a and b) = 1/4 and p(a and not b)=1/3 are p(a and b) and p(a and not b) independent?
@franciscochavez17096 жыл бұрын
I think the author had a mind-slip: IF P(A|B)=P(A) THEN events A & B ARE independent.
@allstarz69z12 жыл бұрын
thk u so much!
@mathteacher26514 жыл бұрын
Well done!
@ccuuttww4 жыл бұрын
thanks
@parthchhatre16565 жыл бұрын
Very helpful
@proshakshin38758 жыл бұрын
you sound like Tobey Maguire
@d.pressley17494 жыл бұрын
Pro Shakshin aha he fors
@d.pressley17494 жыл бұрын
He does* I meant
@vineetmadaan294010 жыл бұрын
good job.....but in case of trivial, both can be possible simultaneously.....
@SuperBrian334412 жыл бұрын
Remember to but your "S" in their...THX ALOT anywyas
@Muchaaachooos8 жыл бұрын
"events that cant happen at the same time" is that P(AUB) or P(AnB)
@mysmallcap7 жыл бұрын
n
@gooddeedsleadto74995 жыл бұрын
Why is pulling red card & spade not independent or dependent? It is not intuitive Yes the probability, given it is spade it is red is different from the probability it is red. why do we call it independent?
@EpicPwnzor133711 жыл бұрын
This didn't help at all, it's like learning probability with a jar of marbles, all over again.
@andybeyond95464 жыл бұрын
If Lil Pump stayed at Harvard
@allisonbenjamin60203 жыл бұрын
Thank you, this was helpful!
@peaches__10 жыл бұрын
excellent!
@potaytopotawto870411 жыл бұрын
"Two events that are independent do not have to be mutually exclusive, or disjoint." You're right that they don't have to be. But you didn't mention that they HAVE TO NOT BE. I mean: "Two events that are independent HAVE TO NOT BE mutually exclusive, or disjoint."
@potaytopotawto870411 жыл бұрын
This video is misleading. For any two events that are both non-trivial, which means that there probabilities aren't 0, independent and disjoint are opposites, just like how yes is the opposite of no and true is the opposite of false. Independent = not disjoint. Disjoint = dependent.
@Aadil2Adnan3 жыл бұрын
This helped, thanks :D
@vivektiwari13534 жыл бұрын
Misleading video.. Ace n red cards can never be independent.. Becoz 2 ace cards also red So if we draw them one by one they will affect each other.. And the way he compared..data..it's totally misleading concepts using mathematical tactics