Discussion Forum

 Let A and B be two events such that    $P(\overline{A\cup B})=\frac{1}{6}, P(A\cap B)=\frac{1}{4}$   and 

$P(\overline{A})=\frac{1}{4}$ , where  $\overline{A}$  stands  for the complement  of the event A. Then , the events A and B are

Answer:

Explanation:

 Given ,   $P(\overline{A\cup B})=\frac{1}{6}, P(A\cap B)=\frac{1}{4}$

  $P(\overline{A})=\frac{1}{4}$

 $\therefore$           $P(A\cup B)=1-P(\overline{A\cup B)}$

                       =$1-\frac{1}{6}=\frac{5}{6}$

 and                   $P(A)=1-P(\overline{A})$

                    =  $1-\frac{1}{4}=\frac{3}{4}$

   $P(A \cup B)=P(A)+P(B)-P(A\cap B)$

 $\Rightarrow$   $\frac{5}{6}=\frac{3}{4}+P(B)-\frac{1}{4}$

$P(B)=\frac{1}{3}\Rightarrow$  A and B are not equally  likely

   $P(A\cap B)= P(A).P(B)=\frac{1}{4}$

 So, events are independent.