Question:
If $\bar{E}$ denote the complement or negation of an even $E$, what is the value of $P(E)+P(\bar{E})$ ?
Solution:
Given: $\bar{E}$ denotes the complement or negation of an event $E$.
TO FIND: $P(E)+P(\bar{E})$
CALCULATION: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is $1 .$
Hence $P(E)+P(E)=1$
Hence $P(E)+P(E)=1$