Calculate ##P(C|A')## in the given probability problem

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Homework Statement
See attached ( textbook question).
Relevant Equations
Understanding of conditional probability
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My interest is on part ##b## only. We know that ##A## and ##B## are independent and not mutually exclusive events therefore,

##P(C)=0.7×0.6=0.42##

##P(C|A')=\dfrac{P(C)-P(A∩C)}{P(A')}=\dfrac{0.42-(0.3×0.42)}{0.7}=\dfrac{0.294}{0.7}=0.42## which is wrong according to textbook solution.

Where is my mistake? cheers.
 

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Revisit what the value of Probability of "A intersect C" is.

Are A and C independent?
 
@scottdave wawawawawawa this was a nice one man! Phew. Seen it...
 

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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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