De Morgan's Laws (in a probability context)
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- เผยแพร่เมื่อ 24 ม.ค. 2018
- A discussion of De Morgan's laws, in the context of basic probability. I illustrate De Morgan's laws using Venn diagrams, describe their meaning in a worded example, and show how they might be useful in a probability calculation.
I will get back to statistics videos in the not-too-distant future. Right now, I'm hammering away on probability for a little while.
I really got over-stimulated by the name of job A
That's my dream job
I was hoping someone commented that, and this was the first I saw 😂😂😂
This one has seemed the most intuitive of this entire series to me
I'm glad you found it intuitive. (Then there is nothing to remember!) De Morgan's Laws are interesting. Students often have trouble with the symbolic representation, but the notion that 'not (this or that)' = 'not this and not that' is pretty straightforward to grasp, especially with simple examples. Cheers.
These videos are very clear and useful. Great job! Now I have to go back and like them each individually...
This was incredibly clear - thanks and great job.
Clearly illustrated and explained. Thank you
Straightforward and easy. Nice video!
This is the literal best explanation on the entire internet
Incredible visualisation !
You really know how to teach! Thank you!
Thank you, this was helpful!
Thank you so much for your videos. I passed my probability class with a good grade because of you! (:
You are very welcome! I'm glad to be of help.
@@jbstatistics come back :(
Study for AP computer science this year, this helps a lot, thank you!
Thank you for this, really got the idea behind it after watching this!
Glad, you're back!!
Thanks! It's good to be back!
Thanks!
good demonstration by using the diagrams. Thank you
Amazing explanation!
Thanks!
Thank you so freaking much....please keep the good work
hands down one of the best explanations ive ever heard.
thank you for the examples
amazing video :) literally so easy and simple to understand , thank you 🙏
thank you thank you, this was so helpful!
can we be friends?
you are so cool. Thank you do much professor
so smooth ...thanks jb
amazing and very clear
Thanks! I'm glad to be of help!
Amazing and simple explanation! Thank you:)
Brilliant video
Awesome
Very clear.Thank you
You are very welcome!
thank you for the animation :)
Brilliant!
We Canadians are the best at everything! Well in this case you are Jb. Thanks man!
Thanks! Happy to be of help!
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Let's take it up a notch, because I have a problem that's more complicated. I have 10 digits, A through J, and from those I generate a 4-digit ordered string such as AAAA or ABCD or AJAE. I want to know the probability that the string has at least one A in it. Call the 4 variables that make up the string w,x,y,z, so the string is wxyz. We are then finding the probability of statement S:="w=A or x=A or y=A or z=A". I am not sure if this is correct, but I believe one way to proceed is to find the easier probability of not S = "w != A and x != A and y != A and z != A". The probability that w != A is (1-pr(w=A))=(1-0.1)=0.9, and the same for x,y, and z. Thus, we multiply the probabilities together so that the probability of not S is (0.9)^4 = 6561/10000, and thus the probability of S is (1-Pr(not S))=(1-6561/10000)=3439/10000. So, the probability that at least one digit is an A is 0.3439. I believe this is correct. I don't know how to put this in the language of sets instead of the language of boolean statements.
Thanks!!!!
Thank you
Thank you Thank you Thank you
(You are very welcome!)^3
@@jbstatistics I see what you did there ^_^
Me feeling like an expert after watching the complete playlist ✅
Glad to be of help!
Thanks a lot！
thank you
Thankyou sir
ty king
أحسنت يا رجل
HELP ! im really confused with this ... Not (A and B) = Not A or Not B but it seems like it will have the same result as Not A and Not B ... what am i missing ?
Not (A or B) = Not A and Not B also seems like it will have the same result as Not A or Not B ...
can anyone give me an example with literal numbers so that i can understand how they actually differ in results ?
im not sure i can answer your quetions. my english is bad so sorry for error or grammar mistake
not(p and q) you need to start with inside bracket.
"and" not means "and" it means the space that interact ,or you can say when they are both true.
space only p but not q is not selected bcs p = true but q =false.
space only q but not p is also not selected bcs q = true but p = false.
the space interact between p and q is selected bcs both true
so not (p and q) means that the space that are not interact, or you can say in either one is not true or both are false.(just like the opposite of p and q)
Not p or not q.
No bracket we just start it from left to right.
"or" is means you can accept both true or either one is false.
not p is the space which besides p so outside p is true and inside p is false.outside q is true and inside q is false.
then the interact space p and q will be false because both are touch the space of inside p and inside q. (both false)
So, the other space that not interact between p and q will be selected because they are accepted by "or"(either one is false or both are true)
Not (p or q)
inside bracket first.
"or" only did not accept space both false
p is true and q is true.
the interact space p and q is both true.
the space only have p is p = true and q = false.(accepted bcs not both false) q same as p
so space inside p and q is all true.
then not (p or q)means the space that not true, also means that the space that p = false and q = false.
so the space we take will be the space outside p and outside q because the space is both false(no p and no q).
Not p and not q
no bracket start from left to right.
not p = inside p is false and outside p is true
not q = inside q is false and outside q is true
"and" means interact and both must be true.
so the space that outside p and also outside q will be selected bcs both are true.
Nice video# Dr. Upasana Taneja
Would recommend calling not independent as dependent 90% of the times and only 10% of the times not independent, as the notion of dependency is way stronger and something big to prove in physics or any other real life events. From what i saw so far was the inverse trend with not independent used 97% of the time, which is helping less setting up intuition. My 2 cents.
You fit!👌
@ 4:26, I totally understand how the probability of the compliment of the union of A & B = 1 - 0.65 = 0.35. However, if we look at it from another perspective then the intersection of A compliment & B compliment = 0.80 × 0.40 = 0.32 not 0.35. Anybody can explain the discrepancy here? Where is there 0.03 difference? Thanks in advance.
The events are not independent here. The probability of the intersection of two events is equal to the product of their probabilities if and only if the events are independent.
@@jbstatistics That makes perfect sense. Thank you so much.
Can the last example be represented using a venn diagram.I can't find the probability of the intersection of A and B.Thanks
The last example asks for the probability of the intersection of the *complements* of A and B. This intersection is the same event as the complement of the union of A and B (By De Morgan's laws, as given in the video). Casually speaking, that's the region outside the circles.
How did you get .05 and 0.45?
Sal Salo you have to subtract that overlap. See it is 0.15 in the middle. Well in order to evenly distribute the probability correctly, you need to subtract 0.15 from both A and B probabilities so it makes sense.
Well done
great video and thank you, but is it possible you made an error= 2:44 Not A and Not B
What do you think is in error? It's highly unlikely there's a problem. "Not A and not B" does not come up at 2:44.
thanks.
give us list of all on the way vids pls...
While I have a big list of videos I want to make, and have a notion of where I'm going with all of this, I'm never sure what's coming next until I sit down to record. So I'm not going to post an "upcoming" type of list. (Things come up, and I might go a different route, and I don't want to pen myself in.)
til now, you've done alot to me, i appreciated.
but to catch what ST is, i've need some basics, geometric&harmonic mean, quartiles, range, iqr, SK,... by your way of teaching.
this guy is funny
At 3:50
Why isn't P(A or B) = P(A) + P(B) but rather P(A\B) + P(A and B) + P(B\A)? And also, why aren't these the same? P(A) + P(B) is .2 + .6 = .8. P(A\B) + P(A and B) + P(B\A) is .05 + .15 + .45 = .65
huh?
P(A or B) = P(A) + P(B) - P(AnB) (the addition rule). It is also true that P(A or B) = P(AnB^c) + P(AnB) + P(A^c n B). Both of those work out to 0.65 in the example in this video.
AnB denotes A and B? Then the or in P(A or B) is an exclusive or?
Yes, I'm using AnB to denote the intersection of A and B. I'm using "or" in the inclusive sense, as nearly everyone does in a study of probability. A or B means A or B or both.
Ok thanks, I understand it now for the most part. One last question though, the probabilities A and B need to be normalized somehow, right? But they do not sum to 1 here. Why is that? I mean if the probabilities for "getting the job" were say .9 for both A and B and the the probability of getting both (P(AnB)) were 0, then P(A or B) would be .9 + .9 - 0... so A and B must be normalized, but if they were normalized they would some to 1, but they don't. Why is that?
So Easy!
Do D'Morgan's laws apply to mutually exclusive events as well?
They are universal laws that have no conditions other than we're discussing two events in a sample space.
@@jbstatistics I have the following question. Let's say C - is the event that a manager is in the office and P(C) = 0.48. And D - the event that the manager is at home with P(D) = 0.27. Find the probability that she will neither be in her office or at home. I have two solutions but they do not agree. Solution 1: C and D are mutually exclusive. P(CvD) ' = 1 - (.48+.27)= .25
Solution 2: P(CvD)' = P(C') and P(D')= P(C')P(D') = (1-0.48)(1-0.27)= (0.52)(0.73)= 0.38
I know solution 2 is incorrect but I can't find the error in my logic. Please help.
@@jbstatistics test
@@Chris-ng9zi The error is that solution 2 implicitly assumes that being in the office is independent of being at home, which of course it is not. The probability of an intersection is equal to the product of the individual probabilities if and only if the events are independent.
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