# Let A and B be two events such that P(AUB)’=1/6. P(AnB)=1/4 and P(A)’=1/4 where A’ stands for complement of event . Then events A and B are

Question : Let A and B be two events such that $P\left( \overline{A\cup B}\right) =\dfrac{1}{6}\cdot P\left( A\cap B\right) =\dfrac{1}{4}andP\left( \overline{A}\right) =\dfrac{1}{4}$ where $\overline{A}$ stands for complement of event . Then events A and B are

(A) Equally likely but not independent

(B) Equally likely and mutually exclusive

(C) Mutually exclusive and independent

(D) Independent but not equally likely

Solution :

We have, $P\left( \overline{A\cup B}\right) =\dfrac{1}{6}\cdot P\left( A\cap B\right) =\dfrac{1}{4}and P\left( \overline{A}\right) =\dfrac{1}{4}$,

$P\left( A\right) =1-P\left( \overline{A}\right) =1-\dfrac{1}{4}=\dfrac{3}{4}$

therefore, $P\left( \overline{A\cup B}\right) =1-P\left( A\cup B\right)$ $P\left( \overline{A\cup B}\right)=1-\left[ P\left( A\right) +P\left( B\right) -P\left( A\cap B\right) \right]$

$\Rightarrow \dfrac{1}{6}=1-\dfrac{3}{4}-P\left( B\right) +\dfrac{1}{4}$

on simplification we get, $P\left( B\right) =1-\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{1}{3}$

Since $P\left( A\cap B\right) =P\left( A\right) P\left( B\right)$ and $P\left( A\right) \neq P\left( B\right)$ therefore A and B are independent but not equally likely.

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