Probability 3 Question 25
25. If $\bar{E}$ and $\bar{F}$ are the complementary events of $E$ and $F$ respectively and if $0<P(F)<1$, then
$(1998,2 M)$
(a) $P(E / F)+P(\bar{E} / F)=1$
(c) $P(\bar{E} / F)+P(E / \bar{F})=1$
(b) $P(E / F)+P(E / \bar{F})=1$
(d) $P(E / \bar{F})+P(\bar{E} / \bar{F})=1$
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Solution:
- (a) $P(E / F)+P(\bar{E} / F)=\frac{P(E \cap F)}{P(F)}+\frac{P(\bar{E} \cap F)}{P(F)}$
$$ \begin{aligned} & =\frac{P(E \cap F)+P(\bar{E} \cap F)}{P(F)} \\ & =\frac{P(F)}{P(F)}=1 \end{aligned} $$
Therefore, option (a) is correct.
(b) $P(E / F)+P(E / \bar{F})=\frac{P(E \cap F)}{P(F)}+\frac{P(E \cap \bar{F})}{P(\bar{F})}$
$$ =\frac{P(E \cap F)}{P(F)}+\frac{P(E \cap \bar{F})}{1-P(F)} \neq 1 $$
Therefore, option (b) is not correct.
(c) $P(\bar{E} / F)+P(E / \bar{F})=\frac{P(\bar{E} \cap F)}{P(F)}+\frac{P(E \cap \bar{F})}{P(\bar{F})}$
$$ =\frac{P(\bar{E} \cap F)}{P(F)}+\frac{P(E \cap \bar{F})}{1-P(F)} \neq 1 $$
Therefore, option (c) is not correct.
(d) $P(E / \bar{F})+P(\bar{E} / \bar{F})=\frac{P(E \cap \bar{F})}{P(\bar{F})}+\frac{P(\bar{E} \cap \bar{F})}{P(\bar{F})}$
$$ \begin{aligned} & =\frac{P(E \cap \bar{F})+P(\bar{E} \cap \bar{F})}{P(\bar{F})} \\ & =\frac{P(\bar{F})}{P(\bar{F})}=1 \end{aligned} $$
Therefore, option (d) is correct.
