If P(A)=23, P(B)=12 and P(A∪B)=56, then events A and B are

A
mutually exclusive
B
independent as well as mutually exclusive
C
independent
D
dependent only on $A$

Solution:

We have,

$P (A) = \frac{2}{3}, P (B) = \frac{1}{2}$ and $P (A \cup B) = \frac{5}{6}$

$\begin{array}{ll} ∴ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ ∴ P (A \cap B) = P (A) + P (B) - P (A \cup B) \\ \Rightarrow P (A \cap B) = \frac{2}{3} + \frac{1}{2} - \frac{5}{6} = \frac{1}{3} \end{array}$

Clearly, $P (A \cap B) = P (A) P (B)$

Hence, $A$ and $B$ are independent events.

Since independent events are never mutually exclusive.

So, option $(c)$ is correct.

If $P (A) = \frac{2}{3}, P (B) = \frac{1}{2}$ and $P (A \cup B) = \frac{5}{6},$ then events $A$ and $B$ are

Solution:

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