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)

= $${3 \over 8} + {1 \over {16}}$$ = $${7 \over {16}}$$