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14.Probability
medium
If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P\,(F) < 1,$ then
A
$E$ and ${F^c}$ (the complement of the event $F$) are independent
B
${E^c}$ and ${F^c}$ are independent
C
$P\,\left( {\frac{E}{F}} \right) + P\,\left( {\frac{{{E^c}}}{{{F^c}}}} \right) = 1$
D
All of the above
(IIT-1989)
Solution
(d) $P(E \cap F) = P(E)\,.\,P(F)$
Now, $P(E \cap {F^c}) = P(E) – P(E \cap F) = P(E)[1 – P(F)] = P(E)\,.P({F^c})$
and $P({E^c} \cap {F^c}) = 1 – P(E \cup F) = 1 – [P(E) + P(F) – P(E \cap F)$
$ = [1 – P(E)][1 – P(F)] = P({E^c})\,P({F^c})$
Also $P(E/F) = P(E)$ and $P({E^c}/{F^c}) = P({E^c})$
$ \Rightarrow P(E/F) + P({E^c}/{F^c}) = 1.$
Standard 11
Mathematics
