steve common said:Is it
( P(Ch Scores) + P(S Scores) ) * 60% = 0.18%
nah, you can't add them unless they're mutually exclusive. Doing it this way, you're
giving Chucky zero chance -- which is a bit mean (-;
You want:
P(C|S) = P(S|C) P(C)
-----------
P(S)
Where P(C) is the probability that Chucky gets lucky, P(C|S) is the conditional
prob Chuck gets lucky given that S does, etc.
So P(C|S) = 0.6 * 0.001 / 0.002 = 0.3 ---> there's a 30% chance Chucky is lucky
(or, if you give yourself a .3% chance, his odds drop to 20%)
Quoted message said:(**) I remember stuff about bags with black balls and white balls all
mixed together - glad this thread didn't take us there sooner :-P
It could be arranged (-;
Suppose there is one bag containing 2 black balls and a white ball, and another
bag containing 2 black balls and 2 white balls. A bag is selected at random and
a ball is pulled out. The ball is black. What is the probability that it came
from the bag that contains one white ball ?
Cheers,
--
Elflord