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14.Probability
hard
If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then
A
$P\,(A \cup B) \ge \frac{2}{3}$
B
$\frac{1}{6} \le P(A \cap B) \le \frac{1}{2}$
C
$\frac{1}{6} \le P(A' \cap B) \le \frac{1}{2}$
D
All of the above
Solution
(d) We have $P(A \cup B) \ge \max .$ $\{ P(A),\,P(B) \} = \frac{2}{3}$
$P(A \cap B) \le \min .$$\{ P(A),P(B)\} = \frac{1}{2}$
and $P(A \cap B) = P(A) + P(B) – P(A \cup B) \ge P(A) – P(B) – 1 = \frac{1}{6}$
$ \Rightarrow \frac{1}{6} \le P(A \cap B) \le \frac{1}{2}$
$P\,(A' \cap B) = P(B) – P(A \cap B)$
Hence $\frac{2}{3} – \frac{1}{2} \le P(A' \cap B) \le \frac{2}{3} – \frac{1}{6}$
$ \Rightarrow \frac{1}{6} \le P(A' \cap B) \le \frac{1}{2}$.
Standard 11
Mathematics
