Let A, B, C be three events. If the probability of occurring exactly one out of A and B is 3 5, exactly one of B and C is 1 5, exactly one of C and A…

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

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

From the given information:

$$P(A) + P(B) - 2P(A \cap B) = \dfrac{3}{5} \quad \cdots (1)$$P(A)+P(B)−2P(A∩B)=53​⋯(1)

$$P(B) + P(C) - 2P(B \cap C) = \dfrac{1}{5} \quad \cdots (2)$$P(B)+P(C)−2P(B∩C)=51​⋯(2)

$$P(C) + P(A) - 2P(C \cap A) = \dfrac{3}{5} \quad \cdots (3)$$P(C)+P(A)−2P(C∩A)=53​⋯(3)

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$$(1) + (2) + (3):$$(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∩B)+P(B∩C)+P(C∩A)]=53​+51​+53​

$$2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = \dfrac{7}{5}$$2[P(A)+P(B)+P(C)]−2[P(A∩B)+P(B∩C)+P(C∩A)]=57​

$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = \dfrac{7}{10}$$P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)=107​

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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∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)

$$P(A \cup B \cup C) = \dfrac{7}{10} + \dfrac{4}{25}$$P(A∪B∪C)=107​+254​

$$= \dfrac{35}{50} + \dfrac{8}{50}$$=5035​+508​

$$= \dfrac{43}{50}$$=5043​

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Therefore, the probability of occurring at least one of them is $\dfrac{43}{50}$5043​.