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14.Probability
hard
यदि $A$ व $B$ दो घटनायें इस प्रकार हैं कि $P(A) = \frac{1}{2}$ व $P(B) = 2/3,$ तो
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
उपरोक्त सभी
Solution
(d) चूँकि $P(A \cup B) \ge $ अधिकतम $\{ P(A),\,P(B) \}= \frac{2}{3}$
$P(A \cap B) \le $ न्यूनतम $\{ P(A),P(B)\} = \frac{1}{2}$
व$P(A \cap B) = P(A) + P(B) – P(A \cup B) \ge P(A) – P(B) – 1 = \frac{1}{6}$
$ \Rightarrow \frac{1}{6} \le P(A \cap B) \le \frac{1}{2}$
$P\,(A' \cap B) = P(B) – P(A \cap B)$
अत: $\frac{2}{3} – \frac{1}{2} \le P(A' \cap B) \le \frac{2}{3} – \frac{1}{6}$
$ \Rightarrow \frac{1}{6} \le P(A' \cap B) \le \frac{1}{2}$.
Standard 11
Mathematics