If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three events of a random experiment with…
If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three events of a random experiment with $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$, then the probability that neither A nor B occurs is
$0.5$
$0.15$
$0.13$
$0.12$
Solution
Given, $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3$
$\begin{aligned}
& P(A \cap B)=0.2 \\
& \text { Now, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow P(A \cup B)=0.4+0.3-0.2=0.5
\end{aligned}$
Since, probability of neither A nor B occurs
$\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-0.5=0.5$