If \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2\) and $P(A \cup B)=0.7$ then…
If \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2\) and $P(A \cup B)=0.7$ then $P^{\prime}\left(A^{\prime}\right)+P\left(B^{\prime}\right)$ is
- $1.1$
- $0.6$
- $1.8$
- $1.6$
Solution
$\begin{aligned} & P\left(A^{\prime}\right)+P\left(B^{\prime}\right)=1-P(A)+1-P(B) \\ & =2-\{P(A)+P(B)\} \\ & =2-\{P(A \cup B)+P(A \cap B)\} \\ & =2-\{0.7+0.2\} \\ & =2-0.9 \\ & =1.1\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 2)
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