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 out of B and C is $\frac{1}{5}$, exactly one out 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

$\frac{43}{50}$

The correct answer is Option (3) → $\frac{43}{50}$

Let $P(A \cap B \cap C) = \frac{4}{25}$

Given: $P(\text{exactly one of A and B}) = P(A \cup B) - 2P(A \cap B) = \frac{3}{5}$ → $P(A \cap \bar{B}) + P(\bar{A} \cap B) = \frac{3}{5}$

$P(\text{exactly one of B and C}) = P(B \cap \bar{C}) + P(\bar{B} \cap C) = \frac{1}{5}$

$P(\text{exactly one of C and A}) = P(C \cap \bar{A}) + P(\bar{C} \cap A) = \frac{3}{5}$

Let $P(A) = x$, $P(B) = y$, $P(C) = z$ and $P(A \cap B \cap C) = t = \frac{4}{25}$

We know: $P(\text{exactly one of A and B}) = x + y - 2P(A \cap B)$ → $P(A \cap B) = \frac{x+y - 3/5}{2}$

Similarly, $P(B \cap C) = \frac{y+z - 1/5}{2}$, $P(C \cap A) = \frac{z+x - 3/5}{2}$

Probability of at least one: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$

Substitute: $P(A \cup B \cup C) = (x+y+z) - \left[\frac{x+y-3/5}{2} + \frac{y+z-1/5}{2} + \frac{z+x-3/5}{2}\right] + 4/25$

Simplify:

$x+y+z - \frac{2(x+y+z) - 7/5}{2} + 4/25 = x+y+z - (x+y+z - 7/10) + 4/25 = 7/10 + 4/25 = 35/50 + 8/50 = 43/50$

Therefore, $P(\text{at least one}) = \frac{43}{50}$