For events A,B,C: P(exactly one of A,B)=P(exactly one of B,C)=P(exactly one of C,A)=14, an…

Question

For events $A,B,C$: $P(	ext{exactly one of }A,B)=P(	ext{exactly one of }B,C)=P(	ext{exactly one of }C,A)=\dfrac14$, and $P(	ext{all three})=\dfrac{1}{16}$. Find $P(	ext{at least one}).$

Accepted Answer

Adding the three "exactly one" equations: $2\sum P(A)-2\sum P(A\cap B)=\dfrac34 \Rightarrow \sum P(A)-\sum P(A\cap B)=\dfrac38$.

$$P(A\cup B\cup C) = \frac38+\frac{1}{16}=\frac{7}{16}$$

Solutions · 1