# Math Help - easy probability question

1. ## easy probability question

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.

2. 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)

3. 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...

4. Originally Posted by 0123
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...

P(R)= P(R|C1) + P(R|C2) makes no sense.

also P(R|C1) + P(R|C2) could be greater than 1...that's a problem.

5. Hello, 0123!

[quote[If I wrote: .P(R) .= .P(R|C1) + P(R|C2)[/tex]
what sense would it have (if any)?
[/quote]

6. Hello, 0123!

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.
You do know that this is wrong, don't you?

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%

7. Oh. Now I know.