Let $A$ and $B$ be the events such that $P(A)=\frac{1}{2}, P(B)=\frac{7}{12}$ and $P(\operatorname{not} A$ or $n o t B)=\frac{1}{4}$.State whether A and B are(i) mutually exclusive(ii) independent

Home » Let and be the events such that and or .State whether A and B are(i) mutually exclusive(ii) independent

Let $A$ and $B$ be the events such that $P(A)=\frac{1}{2}, P(B)=\frac{7}{12}$ and $P(\operatorname{not} A$ or $n o t B)=\frac{1}{4}$ .
State whether A and B are
(i) mutually exclusive
(ii) independent

Let $A$ and $B$ be the events such that $P(A)=\frac{1}{2}, P(B)=\frac{7}{12}$ and $P(\operatorname{not} A$ or $n o t B)=\frac{1}{4}$ .
State whether A and B are
(i) mutually exclusive
(ii) independent

Given: $A$ and $B$ are the events such that $P(A)=\frac{1}{2}$ and $P(B)=\frac{7}{12}$ and
$P(\operatorname{not} A$ or $\operatorname{not} B)=\frac{1}{4}$
To Find: i)if A and B are mutually exclusive
Since $\mathrm{P}($ not $\mathrm{A}$ or $\mathrm{not} \mathrm{B})=\frac{1}{4}$ i.e., $\mathrm{P}(\bar{A} \cup \bar{B})=\frac{1}{4}$
we know that, $P(\bar{A} \cup \bar{B})=P(A \cap B)^{\prime}=1-P(A \cap B)=\frac{1}{4}$
$\Rightarrow P(A \cap B)=1-\frac{1}{4}=\frac{3}{4}$ Equation 1
Since for two mutually exclusive events $P(A \cap B)=0$
But here $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq 0$
Therefore, $A$ and $B$ are not mutually exclusive.
ii) If $A$ and $B$ are independent
The condition for two events to be independent is given by
$P\left(E_{1} \cap E_{2}\right)=P\left(E_{1}\right) \times P\left(E_{2}\right)$
$\begin{array}{l} =\frac{1}{2} \times \frac{7}{12} \\ =\frac{7}{24} \text { Equation } 2 \end{array}$
Since Equation $1 \neq$ Equation 2
$\Rightarrow A$ and $B$ are not independent