If E and F are two events such that P(E) = 0.8, P(F) = 0.7 and P(E \( \cap \) F) = 0.6 then P(\( \overline{\mathrm{E}} \) / \( \overline{\mathrm{F}} \)) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{3} \)
By the definition of conditional probability:
\( \displaystyle P(\overline{E}/\overline{F})=\frac{P(\overline{E}\cap\overline{F})}{P(\overline{F})} \)
By De Morgan's law, \( \overline{E}\cap\overline{F}=\overline{E\cup F} \):
\( \displaystyle P(E\cup F)=0.8+0.7-0.6=0.9 \)
\( \displaystyle P(\overline{E}\cap\overline{F})=1-0.9=0.1 \)
\( \displaystyle P(\overline{F})=1-0.7=0.3 \)
\( \displaystyle P(\overline{E}/\overline{F})=\frac{0.1}{0.3} \)
\( \displaystyle =\frac{1}{3} \)