The probability that at least one of the events $E_{1}$ and $E_{2}$ occurs is 0.6 If the probability of the simultaneous occurrence of $E_{1}$ and $E_{2}$ is 0.2, find $P ({¯ E}_{1}) + P ({¯ E}_{2}) .$

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Solution

Given, $P (E_{1} \cup E_{2}) = 0.6$ and $P (E_{1} \cap E_{2}) = 0.2.$

$∴ P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2})$

$\Rightarrow P (E_{1}) + P (E_{2}) = P (E_{1} \cup E_{2}) + P (E_{1} \cap E_{2}) = (0.6 + 0.2) = 0.8$

$\Rightarrow P (E_{1}) + P (E_{2}) = 0.8$

$\Rightarrow {1 - P ({¯ E}_{1})} + {1 - P ({¯ E}_{2})} = 0.8$

$\Rightarrow P ({¯ E}_{1}) + P ({¯ E}_{2}) = (2 - 0.8) = 1.2$

Hence, $P ({¯ E}_{1}) + P ({¯ E}_{2}) = 1.2.$

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