Occuring Events- Probabilities

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Homework Statement


A box contains two white, two red, and two blue poker chips. Two chips are randomly chosen without replacement, and their colors are noted. Define the following events:
A) Both chips are of the same color
B)Both chips are red
C)At least one chip is red or white
Find P(A/B), P(B/A^c), P(B/C), P(A/C), P(C/A^c)


Homework Equations





The Attempt at a Solution


A) So at the start there are 6 chips, probability of pulling out the first one of any color would be 6/6 then there is one less in the bag but we now need the same color as we hauled out the first time. So if you have a red one out for example there are 5 chips left and only one is red so is it 1/5 chance?
 
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Wouldnt the answer be the same for both A and B ?
 
Do you understand what A and B ask? Neither A nor B asks for a probability. That is the last question in the problem (that is not labeled). A and B merely asks for a lists of the "events"- that is, a list of the pairs of chips that satisfy the conditions "two chips of the same color" for A and "two red chips" for B. No, the answers are not the same- and nothing that you said under "A)" has anything to do with the answer to A.
 
So for the first one it can be defined as {WW,BB,RR} because those are the chips that can satisfy the conditions?
 
Can someone please help me this is the easiest of 4 questions I have left I'm lost
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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