In two events P(A cup B) = 5/6 , P({A^c}) = 5/6 , P(B) = 2/3, then A and B are

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14.Probability

medium

In two events $P(A \cup B) = 5/6$, $P({A^c}) = 5/6$, $P(B) = 2/3,$ then $A$ and $B$ are

A

Independent

B

Mutually exclusive

C

Mutually exhaustive

D

Dependent

Solution

(b) $P(A{^c}) = 1 – P(A)$ $ \Rightarrow $ $P(A) = \frac{1}{6}$

Now $P(A \cup B) = P(A) + P(B) – P(A \cap B)$

$⇒ \frac{5}{6} = \frac{1}{6} + \frac{2}{3} – P(A \cap B)$ ==> $P(A \cap B) = 0$

Hence, events $A$ and $B$ are mutually exclusive.

Standard 11

Mathematics

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