Let A, B, C be three events. If the probability of occurring exactly one out of A and B is $\frac{3}{5}$, exactly one of B and C is $\frac{1}{5}$, exactly one of C and A is $\frac{3}{5}$ and that of occurring of three events is $\frac{4}{25}$, then the probability of occurring at least one of them is

The probability of exactly one of two events $X$ and $Y$ occurring is:

$P(\text{exactly one of } X \text{ and } Y) = P(X) + P(Y) - 2P(X \cap Y)$

From the given information:

$P(A) + P(B) - 2P(A \cap B) = \dfrac{3}{5} \quad \cdots (1)$

$P(B) + P(C) - 2P(B \cap C) = \dfrac{1}{5} \quad \cdots (2)$

$P(C) + P(A) - 2P(C \cap A) = \dfrac{3}{5} \quad \cdots (3)$

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$(1) + (2) + (3):$

$2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = \dfrac{3}{5} + \dfrac{1}{5} + \dfrac{3}{5}$

$2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = \dfrac{7}{5}$

$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = \dfrac{7}{10}$

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Using the union formula:

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

$P(A \cup B \cup C) = \dfrac{7}{10} + \dfrac{4}{25}$

$= \dfrac{35}{50} + \dfrac{8}{50}$

$= \dfrac{43}{50}$

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Therefore, the probability of occurring at least one of them is $\dfrac{43}{50}$.