Have to point out that the dice table is wrong for permutation outcomes because you only counted the 1,1 2,2 ..... only once when they are needed to be counted twice
The diagonal elements have the same likelihood of occurrence as the off-diagonal elements. They are not twice as likely to happen as you suggested. P(LH=1, RH=1) and P(RH=1,LH=1) correspond to a single event in the sample space. We don’t need to count it twice. P(1,1) = P(LH =1) P(RH=1) =(1/6)(1/6)=1/36. Left hand toss and right hand toss are independent experiments.
going through this entire playlist a week before my probability exam >>
literally me
Good explanation !
Great explanation of conditional probability! Thank you!
Your lessons are amazing!
thank you very much, you saved my worthless life. now I have one.
Yes I agreee
Very helpful, keep up the good work mate!
Shouldn't P(AnB) be 3/6 instead of 3/36?
very good job
Now redo with modern animations :p
thanks
10/10
Have to point out that the dice table is wrong for permutation outcomes because you only counted the 1,1 2,2 ..... only once when they are needed to be counted twice
The diagonal elements have the same likelihood of occurrence as the off-diagonal elements. They are not twice as likely to happen as you suggested. P(LH=1, RH=1) and P(RH=1,LH=1) correspond to a single event in the sample space. We don’t need to count it twice. P(1,1) = P(LH =1) P(RH=1) =(1/6)(1/6)=1/36. Left hand toss and right hand toss are independent experiments.
Actuarial Path thank you very much for responding back and explaining, btw great videos
@@StatCourses pls i didn't under the A intersection B