The probability of at least one of the events | vedclass

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(C) Given that the probability of at least one event occurring is $P(A \cup B) = 0.6$. By De Morgan's Law,$P(A \cup B) = 1 - P(A' \cap B') = 0.6$,so $

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$1.1$

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(C) Given that the probability of at least one event occurring is $P(A \cup B) = 0.6$. By De Morgan's Law,$P(A \cup B) = 1 - P(A' \cap B') = 0.6$,so $P(A' \cap B') = 0.4$. We know that $P(A \cap B) = 0.3$. Using the formula $P(A' \cup B') = P(A') + P(B') - P(A' \cap B')$,where $P(A' \cup B') = 1 - P(A \cap B) = 1 - 0.3 = 0.7$. Substituting the values: $0.7 = P(A') + P(B') - 0.4$. Therefore,$P(A') + P(B') = 0.7 + 0.4 = 1.1$.

The probability of at least one of the events $A$ and $B$ occurring is $0.6$. If the probability of both $A$ and $B$ occurring simultaneously is $0.3$,then $P(A') + P(B') = \dots$

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