Question

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

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

Answer 1

The correct option is B $\frac{7}{16}$

$P\left(A\right)+P\left(B\right)–2P(A\cap B)=\frac{1}{4}P\left(B\right)+P\left(C\right)–2P(B\cap C)=\frac{1}{4}P\left(A\right)+P\left(C\right)–2P(A\cap C)=\frac{1}{4}P(A\cap B\cap C)=\frac{1}{16}\therefore P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)–P(B\cap C)–P(A\cap C)+P(A\cap B\cap C)=\frac{7}{16}$