>>810697>how would you answer this problem, 1/3 or 100%? It all boils down to how we personally interpret the wording of the problem in this thread, how we mathematically deal with fact that we are told we already picked a green ball.No it doesn't, that's just you not being able to calculate conditional probablility.
>you pick a ball // establish probability distribution1 (g) (1)
2 (r) (1)
3 (r) (1)
>it's green // impose filter1 (g) (1) IN
2 (r) (1) OUT
3 (r) (1) OUT
//Discard filtered lines and sum up probability
1 (g) (1) IN
ball is green = (1)/(1)
>probability of it being green is (1)/(1)=100%// answer question
>what is the probability that you picked a green ball100%
Let's work out the initial problem too:
>you pick a random box then you pick a random ball // establish distribution1,1,it's green, other ball is green,(1)
1,2,it's green, other ball is green,(1)
2,1,it's red, other ball is green,(1)
2,2,it's green, other ball is red,(1)
3,1,it's red, other ball is red,(1)
3,2,it's red, other ball is red,(1)
>it [the first ball] is green //impose filter1,1,it's green, other ball is green,(1) IN
1,2,it's green, other ball is green,(1 IN)
2,1,it's red, other ball is green,(1) OUT
2,2,it's green, other ball is red,(1) IN
3,1,it's red, other ball is red,(1) OUT
3,2,it's red, other ball is red,(1) OUT
//Discard filtered lines and sum up probability
1,1,it's green, other ball is green,(1) IN
1,2,it's green, other ball is green,(1 IN)
2,2,it's green, other ball is red,(1) IN
other ball is green (1)+(1)/(1)+(1)+(1) = 2/3
other ball is red (1)+(1)/(1)+(1)+(1) = 1/3
// answer question
>probability the other ball is green is 2/3It's really simple, if you're being asked a question where the conditional is in the past tense, you're being asked about the situation after the "apply filter" step.
