Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below:

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}{E}_1\text{and}{E}_2\text{be mutually exclusive events, then}& 1& E_1\cap E_2=E_1\\ \text{(B)}& \text{If}{E}_1\text{and}{E}_2\text{are mutually exclusive and exhaustive events, then}& 2& (E_1-E_2)\cup (E_1\cap E_2)=E_1\\ \text{(C)}& \text{If}{E}_1\text{and}{E}_2\text{have common outcomes, then}& 3& E_1\cap E_2=\varphi ,E_1\cup E_2=S\\ \text{(D)}& \text{If}{E}_1\text{and}{E}_2\text{are events such that}{E}_1\subset E_2\text{, then}& 4& E_1\cap E_2=\varphi \end{array}$

Which of the following is the only CORRECT combination?

• A. $A\to 1,B\to 3,C\to 2,D\to 4$
• B. $A\to 4,B\to 3,C\to 2,D\to 1$
• C. $A\to 4,B\to 2,C\to 3,D\to 1$
• D. $A\to 3,B\to 4,C\to 1,D\to 2$

Answer 1

The correct option is B $A\to 4,B\to 3,C\to 2,D\to 1$

$A)\text{If}{E}_1\text{and}{E}_2\text{be mutually exclusive events}$
$\Rightarrow E_1\cap E_2=\varphi$

$\left(B\right)$ If $E_1$ and $E_2$ are mutually exclusive and exhaustive events
$\Rightarrow E_1\cap E_2=\varphi ,E_1\cup E_2=S$

$\left(C\right)$ If $E_1$ and $E_2$ have common outcomes $\Rightarrow (E_1-E_2)\cup (E_1\cap E_2)=E_1$

$\left(D\right)$ If $E_1$ and $E_2$ are events such that $E_1\subset E_2\Rightarrow E_1\cap E_2=E_1$

So, correct combination is $A\to 4,B\to 3,C\to 2,D\to 1$