P(R)=P[(R and C1) or (R and C2)]
=P(R and C1) + P(R and C2)
=P(R|C1)*P(C1)+P(R|C2)*P(C2)
Hello! Probability is troubling me. excuse me for the stupid thing I might be asking..
So, let's say there are two containers, one with 3 red balls and 7 black balls; the otehr with 6 rad balls and 4 black balls. If I am asked to compute the probabiulity to pick a red ball, intuitively I would sum all the red balls and divide by the total number of balls. Using the formula, I should say P(R)= P(R /\ C1) + P(R/\C2) where C stands for container and /\ intersect... ok but why should I do in this way? I mean, P(R given C1) + P(R given C2) is not correct, but I see it is not correct with numbers, because otherwise I would interpret it as the sum of the probabilities that I pick a red ball from container 1 plus the prob I pick a red ball from container two. But the second formula evidently does not mean this...so what does it mean? Does it have any sense?
Thanks a lot for you help.
Yeah, i understood how to proceed with tyhe formulas wich show the differenr result, but it's the logic(in word I mean)that I don't get. I mean, if I wrote P(R)= P(R|C1) + P(R|C2) which sense would have it had(if any)?
sorry, I'm perhaps creating so much confusion...
Hello, 0123!
You do know that this is wrong, don't you?There are two containers, one with 3 red balls and 7 black balls;
the other with 6 red balls and 4 black balls.
If I am asked to compute the probabiulity to pick a red ball,
intuitively I would sum all the red balls and divide by the total number of balls.
Suppose we have 50 red balls and 50 black balls in two containers.
You pick a container at random and draw a ball from that container.
. . What is the probability that the ball is red?
Answer: It depends on how the balls are distributed.
If Urn #1 had: .1 red ball
and Urn #2 had: .49 red balls and 50 black balls.
Then: .P(red) .= .½(1) + ½(49/99) .≈ .74.7%