Question

For three events $A,B$ and $C$, $P$ (Exactly one of $A$ or $B$ occurs) $=P$ (Exactly one of $B$ or $C$ occurs) $=P$(Exactly one of $C$ or $A$ occurs) $=\frac{1}{4}$and $P$ (All the three events occur simultaneously)$=\frac{1}{16}$. Then the probability that at least one of the events occurs, is.

• A. $\frac{7}{32}$
• B. $\frac{7}{16}$
• C. $\frac{7}{64}$
• D. $\frac{3}{16}$

Answer 1

Answer: (B)

$\frac{7}{16}$

$P$(exactly one of $A$ or $B$)$=P(A\cup B)-P(A\cap B)=\frac{1}{4}=P\left(A\right)+P\left(B\right)-2P(A\cap B)$

$P$(exactly one of $C$ or $B$)$=P(C\cup B)-P(C\cap B)=\frac{1}{4}=P\left(C\right)+P\left(B\right)-2P(C\cap B)$

$P$(exactly one of $A$ or $C$)$=P(A\cup C)-P(A\cap C)=\frac{1}{4}=P\left(A\right)+P\left(C\right)-2P(A\cap C)$

Adding all

$2P\left(A\right)+2P\left(B\right)+2P\left(c\right)-2P(A\cap B)-2P(A\cap C)-2P(B\cap C)=\frac{3}{4}$

$P\left(A\right)+P\left(B\right)+P\left(c\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)=\frac{3}{8}$

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

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