If P(A_1 cup A_2) = 1 - P(A_1^c) P(A_2^c),whe | vedclass

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(B) We are given the condition $P(A_1 \cup A_2) = 1 - P(A_1^c) P(A_2^c)$.Using the complement rule,$P(A^c) = 1 - P(A)$,we can rewrite the expression a

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(B) We are given the condition $P(A_1 \cup A_2) = 1 - P(A_1^c) P(A_2^c)$.Using the complement rule,$P(A^c) = 1 - P(A)$,we can rewrite the expression as:$P(A_1 \cup A_2) = 1 - (1 - P(A_1))(1 - P(A_2))$Expanding the right side:$P(A_1 \cup A_2) = 1 - (1 - P(A_1) - P(A_2) + P(A_1)P(A_2))$$P(A_1 \cup A_2) = 1 - 1 + P(A_1) + P(A_2) - P(A_1)P(A_2)$$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1)P(A_2)$By the definition of the union of two events,$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)$.Comparing the two expressions,we get $P(A_1 \cap A_2) = P(A_1)P(A_2)$.This is the condition for two events to be independent.

If $P(A_1 \cup A_2) = 1 - P(A_1^c) P(A_2^c)$,where $c$ stands for complement,then the events $A_1$ and $A_2$ are

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