Prove that if $E$ and $F$ are independent events,then so are the events $E$ and $F^{\prime}$.
(N/A) Since $E$ and $F$ are independent,we have:
$P(E \cap F) = P(E) \cdot P(F)$ ......... $(1)$
From the Venn diagram,it is clear that $E \cap F$ and $E \cap F^{\prime}$ are mutually exclusive events and also $E = (E \cap F) \cup (E \cap F^{\prime})$.
Therefore,$P(E) = P(E \cap F) + P(E \cap F^{\prime})$.
Or,$P(E \cap F^{\prime}) = P(E) - P(E \cap F)$.
Substituting from $(1)$:
$P(E \cap F^{\prime}) = P(E) - P(E) \cdot P(F)$
$= P(E) \cdot (1 - P(F))$
$= P(E) \cdot P(F^{\prime})$
Hence,$E$ and $F^{\prime}$ are independent.
$P(E \cap F) = P(E) \cdot P(F)$ ......... $(1)$
From the Venn diagram,it is clear that $E \cap F$ and $E \cap F^{\prime}$ are mutually exclusive events and also $E = (E \cap F) \cup (E \cap F^{\prime})$.
Therefore,$P(E) = P(E \cap F) + P(E \cap F^{\prime})$.
Or,$P(E \cap F^{\prime}) = P(E) - P(E \cap F)$.
Substituting from $(1)$:
$P(E \cap F^{\prime}) = P(E) - P(E) \cdot P(F)$
$= P(E) \cdot (1 - P(F))$
$= P(E) \cdot P(F^{\prime})$
Hence,$E$ and $F^{\prime}$ are independent.
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