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If $A$ and $B$ are two independent events, then the probability of occurrence of at least one of $\mathrm{A}$ and $\mathrm{B}$ is given by $1 -\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$
Solution
We have
$P($ at least one of $A $ and $ B)=P(A \cup B)$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})[1-\mathrm{P}(\mathrm{A})]$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) . \mathrm{P}\left(\mathrm{A}^{\prime}\right)$
$=1-\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \mathrm{P}\left(\mathrm{A}^{\prime}\right)$
$=1-P\left(A^{\prime}\right)[1-P(B)]$
$=1-P\left(A^{\prime}\right) P\left(B^{\prime}\right)$
