KCET · Maths · Probability
If \(A\) and \(B\) are two events such that \(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{2}\) and \(P(A \mid B)=\frac{1}{4}\), then \(P\left(A^{\prime} \cap B^{\prime}\right)\) is
- A \(\frac{1}{4}\)
- B \(\frac{3}{16}\)
- C \(\frac{1}{12}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
Given, \(P(A)=\frac{1}{2}, P(B)=\frac{1}{2}\) and \(P(A \mid B)=\frac{1}{4}\)
\[
\begin{gathered}
P(A \mid B)=\frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{4}=\frac{P(A \cap B)}{1 / 2} \\
\Rightarrow P(A \cap B)=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{gathered}
\]
\[
\begin{aligned}
& \text { Now, } P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B) \\
& =1-[P(A)+P(B)-P(A \cap B)]=1-\left[\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\right] \\
& =1-1+\frac{1}{8}=\frac{1}{8}
\end{aligned}
\]
\[
\begin{gathered}
P(A \mid B)=\frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{4}=\frac{P(A \cap B)}{1 / 2} \\
\Rightarrow P(A \cap B)=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{gathered}
\]
\[
\begin{aligned}
& \text { Now, } P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B) \\
& =1-[P(A)+P(B)-P(A \cap B)]=1-\left[\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\right] \\
& =1-1+\frac{1}{8}=\frac{1}{8}
\end{aligned}
\]
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