Q238 / 270 · NDA 2020 · Mathematics
If $P(A \cup B)=\frac{5}{6}, P(A \cap B)=\frac{1}{3}$ and $P(\operatorname{not} A)=\frac{1}{2}$, then which one of the following is not correct?
- A$P(B)=\frac{2}{3}$
- B$P(A \cap B)=P(B)$
- Correct answerC$P(A \cup B) \gt P(A)+P(B)$
- D$P(\operatorname{not} A$ and $\operatorname{not} B)=P(\operatorname{not} A) P(\operatorname{not} B)$
Solution
$\begin{aligned}
& P(A \cup B)=\frac{5}{6}, P(A \cap B)=\frac{1}{3}, P\left(A^{\prime}\right)=\frac{1}{2} \\
& P(A)=1-P\left(A^{\prime}\right)=1-\frac{1}{2}=\frac{1}{2} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B)
\end{aligned}$
$\frac{5}{6}=\frac{1}{2}+P(B)-\frac{1}{3}$
(a) $P(B)=\frac{5}{6}-\frac{1}{2}+\frac{1}{3}=\frac{5-3+2}{6}=\frac{4}{6}=\frac{2}{3}$
(b) $P(A \cap B)=\frac{1}{3}=\frac{1}{2} \cdot \frac{2}{3}=P(A) \cdot P(B)$
$\therefore P(A \cap B)=P(A) \cdot P(B)$
(c) $P(A \cup B)=\frac{5}{6}$
$\begin{aligned}
& P(A)+P(B)=\frac{1}{2}+\frac{2}{3}=\frac{7}{6} \\
& \therefore P(A \cup B) \lt P(A)+P(B)
\end{aligned}$
Hence, option (c) is not correct.
(d) $P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)$
$\begin{aligned} & 1-P(A \cup B)=1-\frac{5}{6}=\frac{1}{6} \\ & \text {and } P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)=(1-P(A)) \cdot(1-P(B)) \\ & =\left(1-\frac{1}{2}\right) \cdot\left(1-\frac{2}{3}\right)=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}\end{aligned}$
Hence, $P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)$.
Hence, the correct answer is option: 'C'.