Question

Events $A$ and $B$ are such that
$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{7}{12}\text{and}P\left(\text{not A or not B}\right)=\frac{1}{4}$
State whether $A$ and $B$ are independent

Answer 1

Given,

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{7}{12}$

and $P\left(\text{not A or not B}\right)=\frac{1}{4}$

$\Rightarrow P(A^{\prime }\cup B^{\prime })=\frac{1}{4}$

$P(A\cap B{)}^{\prime }=\frac{1}{4}$ $[\because (A^{\prime }\cup B^{\prime })=(A\cap B{)}^{\prime }]$

We know that

$P(A\cap B)=1-P(A\cap B{)}^{\prime }...(1)$

$[\because (A\cap B{)}^{\prime }=1-(A\cap B{)}^{\prime }]$

Putting $P(A\cap B{)}^{\prime }\text{in}(1)$

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

$\therefore P(A\cap B)=\frac{3}{4}$

By the properties of independent events

$P(A\cap B)=P\left(A\right)\times P\left(B\right)...\left(2\right)$

Putting the value of $P\left(A\right),P\left(B\right)\&P(A\cap B)\text{in}\left(2\right)$

$\Rightarrow \frac{3}{4}=\frac{1}{2}\times \frac{7}{12}.$

$\Rightarrow \frac{3}{4}\ne \frac{7}{24}$

$L.H.S.\ne R.H.S.$

Therefore , $A\&B$ are not independent events.