Without condition would be more technically correct term but yes
Without condition would be more technically correct term but yes
A/100×B=A×B/100
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
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.
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That is a lot of money. Any idea how many they are?