kernelPanic

joined 4 years ago
[–] kernelPanic@lemmy.ml 2 points 2 years ago

Work fine for me on plasma wayland. Even the system settings has an option for it

[–] kernelPanic@lemmy.ml 1 points 2 years ago

Wht is going on?

[–] kernelPanic@lemmy.ml 1 points 2 years ago

Without condition would be more technically correct term but yes

[–] kernelPanic@lemmy.ml 2 points 2 years ago

U can also try !sp

[–] kernelPanic@lemmy.ml 1 points 2 years ago

A/100×B=A×B/100

[–] kernelPanic@lemmy.ml 1 points 2 years ago (2 children)

Just count the number of possibilities. If you change there there two possible first choices to win + if you do not change 1 possible choice to win = 3. If you change there is one possible first choice to lose + if you do not change there two possible first choices to lose=3 P(x1)=P(x1') = 3/6

[–] kernelPanic@lemmy.ml 1 points 2 years ago* (last edited 2 years ago) (4 children)

First, fuck you! I couldn't sleep. The possibility to win the car when you change is the possibility of your first choice to be goat, which is 2/3, because you only win when your first choice is goat when you always change.

x1: you win

x2: you change

x3: you pick goat at first choice

P(x1|x2,x3)=1 P(x1)=1/2 P(x3)=2/3 P(x2)=1/2

P(x1|x2) =?

Chain theory of probability:

P(x1,x2,x3)=P(x3|x1,x2)P(x1|x2)P(x2)=P(x1|x2,x3)P(x2|x3)P(x3)

From Bayes theorem: P(x3|x1,x2)= P(x1|x2,x3)P(x2)/P(x1) =1

x2 and x3 are independent P(x2|x3)=P(x2)

P(x1| x2)=P(x3)=2/3 P(x2|x1)=P(x1|x2)P(x2)/P(X1)=P(x1|x2)

P(x1=1|x2=0) = 1- P(x1=1|x2=1) = 1\3 is the probability to win if u do not change.

[–] kernelPanic@lemmy.ml 2 points 2 years ago

The link is so slow

[–] kernelPanic@lemmy.ml 1 points 2 years ago (1 children)

Does it have system tray for gnu linux?

view more: ‹ prev next ›