If E and F are events such that and P(E and F) find : (i) P(E or F) (ii) P(not E and not F)

Engineering

Mathematics

Conditional Probability

Question

If E and F are events such that $P (E) = \frac{1}{4}, P (F) = \frac{1}{2}$ and P(E and F) $= \frac{1}{8}$ find : (i) P(E or F) (ii) P(not E and not F)

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We have given, $P (E) = \frac{1}{4}, P (F) = \frac{1}{2}$ and P(E and F) $= \frac{1}{8}$
(i) We know that P(E or F) = P(E) + P(F) – P(E and F)
∴ P (E or F) $= \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2 + 4 - 1}{8} = \frac{5}{8}$
(ii) From (i), P(E or F) $= P (E \cup F) = \frac{5}{8}$
We have (E∪F)′ = (E′∩F′) [By De Morgan's law]

∴ P(E′∩F′) = P(E∪F)′
Now P(E∪F)′ = 1− P(E∪F) $= 1 - \frac{5}{8} = \frac{3}{8}$
$∴ P (E^{'} \cap F^{'}) = \frac{3}{8}$
Thus P(not E and not F) $= \frac{3}{8}$

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