If A and B are two events such that P(A) = 12 and P(B) = 23, then

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If A and B are two events such that P(A) =

\frac{1}{2}

and P(B) =

\frac{2}{3}

, then

Options:

A . P(A∪B)≥23

B . 16≤P(A′∩B)≤12

C . 16≤P(A∩B)≤12

D . All of the above

Answer: Option D
:
D
We have

P (A \cup B) \geq

max.

{P (A), P (B)} = \frac{2}{3}

P (A \cap B) \leq

min.

{P (A), P (B)} = \frac{1}{2}

and

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

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

P (A^{'} \cap B) = P (B) - P (A \cap 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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