Prob something occurs given something else occurs

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The discussion revolves around calculating conditional probabilities using a scenario with three boxes containing red and black balls. The first part focuses on determining the probability that the first ball drawn from box A is red, given that the second ball drawn from box B is red. The second part shifts to analyzing the duration of telephone calls related to two products, questioning whether to combine the mean and standard deviation to find the probability of a call lasting less than seven minutes. Participants emphasize the importance of identifying events and their intersections in probability calculations. The conversation highlights the need for clarity in applying conditional probability concepts.
Maybe_Memorie
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So suppose we have 3 boxes, A,B,C. Black and Red balls in each.
In A, 3R and 7B.
B, 6R and 4B
C, 8R and 2B

First you take a ball from A. If its R, take one from B. If its B, take one from C.

Given the second is R, what is the prob the first is R?

So I have a tree diagram, but I'm quite unsure about what to do


Also, let's say we have the length of time of telephone calls associated with 2 products.
Mean and standard deviation given.
If you want to find the prob that a random call will take less than 7 minutes, are you supposed to add the mean and SDs together? then proceed as usual?
 
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Here's a link to the wikipedia page on conditional probabilityhttp://en.wikipedia.org/wiki/Conditional_probability#Definition

Have you identified (the probabilities of) the different events and their intersection?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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