Official solution
Given, P (A \[ \cap \] B \[ \cap \] C) = \[{1 \over {16}}\]
P (exactly one of A or B occurs)
= P(A) + P (B) -- 2P (A \[ \cap \] B) = \[{1 \over 4}\] .....(1)
P (Exactly one of B or C occurs)
= P(B) + P (C) -- 2P (B \[ \cap \] C) = \[{1 \over 4}\] .....(2)
P (Exactly one of C or A occurs)
= P(C) + P(A) -- 2P (C \[ \cap \] A) = \[{1 \over 4}\] .....(3)
Adding (1), (2) and (3),we get
2[ P(A) + P(B) + P (C) - P (A \[ \cap \] B)
- P (B \[ \cap \] C) - P (C \[ \cap \] A)] = \[{3 \over 4}\]
\[ \Rightarrow \] P(A) + P(B) + P (C) - P (A \[ \cap \] B)
- P (B \[ \cap \] C) - P (C \[ \cap \] A) = \[{3 \over 8}\]
\[ \therefore \] P(atleast one event occurs)
= 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)
= \[{3 \over 8} + {1 \over {16}}\] = \[{7 \over {16}}\]