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If \( A \) and \( B \) are two events of a sample space \( S \) such that \( P(A)=0.2, P(B)=0.6 \) and \( P(A \mid B)= \)
\( 0.5 \) then \( P\left(A^{\prime} \mid B\right)= \)
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\( 0.5 \) then \( P\left(A^{\prime} \mid B\right)= \)
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2931 Upvotes
Verified Answer
The correct answer is:
\( \frac{1}{2} \)
(D)
\[
\begin{array}{l}
P(A)=0.2, P(B)=0.6 \\
\Rightarrow P(A \mid B)=0.5 \\
\frac{P(A \cap B)}{P(B)}=0.5 \\
P(A \cap B)=0.5(P(B))=(0.5)(0.6)=0.3 \\
\Rightarrow P\left(A^{\prime} \mid B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}=\frac{P(B)-P(A \cap B)}{P(B)} \\
=\frac{0.6-0.3}{0.6}=\frac{0.3}{0.6}=\frac{3}{6}=\frac{1}{2}
\end{array}
\]
\[
\begin{array}{l}
P(A)=0.2, P(B)=0.6 \\
\Rightarrow P(A \mid B)=0.5 \\
\frac{P(A \cap B)}{P(B)}=0.5 \\
P(A \cap B)=0.5(P(B))=(0.5)(0.6)=0.3 \\
\Rightarrow P\left(A^{\prime} \mid B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}=\frac{P(B)-P(A \cap B)}{P(B)} \\
=\frac{0.6-0.3}{0.6}=\frac{0.3}{0.6}=\frac{3}{6}=\frac{1}{2}
\end{array}
\]
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