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(C) Let $P(A) = x, P(B) = y, P(C) = z$. Since $A, B, C$ are independent,$A^c, B^c, C^c$ are also independent.Given $P(A \cap B^{c} \cap C^{c}) = P(A)P

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$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

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(C) Let $P(A) = x, P(B) = y, P(C) = z$. Since $A, B, C$ are independent,$A^c, B^c, C^c$ are also independent.Given $P(A \cap B^{c} \cap C^{c}) = P(A)P(B^c)P(C^c) = x(1-y)(1-z) = \frac{1}{4} \dots (i)$Given $P(A^{c} \cap B \cap C^{c}) = P(A^c)P(B)P(C^c) = (1-x)y(1-z) = \frac{1}{8} \dots (ii)$Given $P(A^{c} \cap B^{c} \cap C^{c}) = P(A^c)P(B^c)P(C^c) = (1-x)(1-y)(1-z) = \frac{1}{4} \dots (iii)$Dividing $(i)$ by $(iii)$: $\frac{x(1-y)(1-z)}{(1-x)(1-y)(1-z)} = \frac{1/4}{1/4} \Rightarrow \frac{x}{1-x} = 1 \Rightarrow x = 1-x \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$.Dividing $(ii)$ by $(iii)$: $\frac{(1-x)y(1-z)}{(1-x)(1-y)(1-z)} = \frac{1/8}{1/4} \Rightarrow \frac{y}{1-y} = \frac{1}{2} \Rightarrow 2y = 1-y \Rightarrow 3y = 1 \Rightarrow y = \frac{1}{3}$.Substituting $x = \frac{1}{2}$ and $y = \frac{1}{3}$ into $(i)$: $\frac{1}{2} 	imes (1 - \frac{1}{3}) 	imes (1-z) = \frac{1}{4} \Rightarrow \frac{1}{2} 	imes \frac{2}{3} 	imes (1-z) = \frac{1}{4} \Rightarrow \frac{1}{3}(1-z) = \frac{1}{4} \Rightarrow 1-z = \frac{3}{4} \Rightarrow z = 1 - \frac{3}{4} = \frac{1}{4}$.Thus,$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.

If $A, B$ and $C$ are three independent events of a random experiment such that $P(A \cap B^{c} \cap C^{c}) = \frac{1}{4}$,$P(A^{c} \cap B \cap C^{c}) = \frac{1}{8}$ and $P(A^{c} \cap B^{c} \cap C^{c}) = \frac{1}{4}$,then $P(A), P(B)$ and $P(C)$ are respectively

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