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14.Probability
easy
यदि $A$ तथा $B$ दो ऐसी घटनाएँ हों कि $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ तथा $P\,(\bar B) = \frac{1}{3},$ तो $P\,(A) = $
A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$\frac{2}{3}$
Solution
(c) $P(A) = P(A \cap B) + P(A \cup B) – P(B)$
$ = \frac{1}{3} + \frac{5}{6} – \frac{2}{3} = \frac{3}{6} = \frac{1}{2}.$
Standard 11
Mathematics