Three events A , B and C are such that A and B are disjoint, A and...

Let $P(A) = x$.

From the given relation $4P(A) = 2P(B) = P(C)$, we have $P(B) = 2x$ and $P(C) = 4x$.

Since $A$ and $B$ are disjoint, $P(A \cap B) = 0$ and $P(A \cap B \cap C) = 0$.

Since $A$ and $C$ are independent, $P(A \cap C) = P(A)P(C) = x(4x) = 4x^2$.

Since $B$ and $C$ are independent, $P(B \cap C) = P(B)P(C) = 2x(4x) = 8x^2$.

Using the formula for the union of three events:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$

Substituting the given values:

$5x = x + 2x + 4x - 0 - 8x^2 - 4x^2 + 0$

$5x = 7x - 12x^2$

$12x^2 - 2x = 0$

$2x(6x - 1) = 0$

Since $x \neq 0$, we get $x = \dfrac{1}{6}$.

Therefore, $P(C) = 4x = 4 \left(\dfrac{1}{6}\right) = \dfrac{2}{3}$.

Answer: $\dfrac{2}{3}$