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

Answer 1

The correct option is A

$\frac{7}{16}$

The explanation for the correct options:

Step1. Expressing the given data:

Given that

$P$ (Exactly one of $A$or $B$occurs) $=P(A\cup B)-P(A\cap B)=\frac{1}{4}$

Similarly

$P$(Exactly one of $B$or $C$occurs) $=P(B\cup C)-P(B\cap C)=\frac{1}{4}$

$P$(Exactly one of $A$or $C$occurs) $=P(A\cup C)-P(A\cap C)=\frac{1}{4}$

Step2. Find the probability that at least one of the events occurs:

We know that

$\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}\mathbf{-}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}\mathbf{=}\mathbf{}\mathit{P}\mathbf{}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{P}\mathbf{}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathbf{–}\mathbf{}\mathbf{2}\mathit{P}\mathbf{}\mathbf{(}\mathit{A}\mathbf{\cap }\mathbf{}\mathit{B}\mathbf{)}\mathbf{}$

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

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

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

Adding all the above equations, we get

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

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

Since, $P$(All the three events occur simultaneously) $=\frac{1}{16}$

$\Rightarrow$ $P(A\cap B\cap C)=\frac{1}{16}$

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

So, $P(A\cup B\cup C)$$=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$

Hence, Option(A) is the correct answer.