If A and B are two events such that , , then

Engineering

Mathematics

Conditional Probability

Question

If A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{2}{3}$ , then

$P (A \cup B) \geq \frac{2}{3}$

$P (A \cup B^{'}) \leq \frac{1}{3}$

$\frac{1}{6} \leq P (A \cap B) \leq \frac{1}{2}$

$\frac{1}{6} \leq P (A^{'} \cap B) \leq \frac{1}{2}$

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We have

P(A∪B)≥max ${P (A) . P (B)} = \frac{2}{3}$

Next $P (A \cap B) = P (A) + P (B) - P (A \cup B) \geq P (A) + P (B) - 1 = \frac{1}{6}$

and P(A∩B)≤min ${P (A) . P (B)} = \frac{1}{2}$

$\Rightarrow \frac{1}{6} \leq P (A \cap B) \leq \frac{1}{2}$

also $P (A \cap B^{'}) = P (A) - P (A \cap B) \leq \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$

P(A′∩B) = P(B) − P(A∩B)
hence $\frac{2}{3} - \frac{1}{2} \leq P (A^{'} \cap B) \leq \frac{2}{3} - \frac{1}{6}$

$\Rightarrow \frac{1}{6} \leq P (A^{'} \cap B) \leq \frac{1}{2}$

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