Question

For two events A & B, which of the following is/are correct?
$\begin{align}
  & i)\text{ (A}\cup \text{B)}\cap \text{B=B} \\
 & ii)\text{ (A}\cup \text{B}{{\text{)}}^{C}}\cap {{\text{B}}^{C}}\text{=(A}\cup \text{B}{{\text{)}}^{C}} \\
 & iii)\text{ }{{\text{A}}^{C}}\cap (A\cap B)\cap {{B}^{C}}=\phi \\
 & iv)\text{ (}{{\text{A}}^{C}}\cap B)\cap (A\cup B)=B \\
\end{align}$

Answer 1

Hint: In this question, we have to find which of the options are correct. As the options consist of the operations of the set, therefore we will be using the set operations to get the solution. We know that the union $(\cup )$ between two sets means the clubbing of both sets and intersection$(\cap )$ between two sets means the common elements of the sets. Thus, we start solving this problem by proving all the four options one-by-one and thus get the required result for the problem.

Complete step by step solution:
According to the question, we have to find which of the given options is/are correct.
Thus, we will apply the set operation to get the required result for the problem.
Option (1) states that $i)\text{ (A}\cup \text{B)}\cap \text{B=B}$
So, we will first solve the left-hand side of the above equation, which is
$\text{ (A}\cup \text{B)}\cap \text{B}$
Thus, we will apply the distributive property $(a\cup b)\cap c=(a\cap c)\cup (b\cap c)$ in the above equation, we get
$\text{ (A}\cap B\text{)}\cup \text{(B}\cap B\text{)}$
As we know that $\text{B}\cap B=B$ , therefore we get
$\text{ (A}\cap B\text{)}\cup B=B=RHS$
Hence it is the correct option.
Option (2) states that $ii)\text{ (A}\cup \text{B}{{\text{)}}^{C}}\cap {{\text{B}}^{C}}\text{=(A}\cup \text{B}{{\text{)}}^{C}}$
So, we will first solve the left-hand side of the above equation, which is
${{\text{(A}\cup \text{B)}}^{C}}\cap {{\text{B}}^{C}}$
Now, we will first apply de morgan’s law ${{(A\cup B)}^{C}}={{A}^{C}}\cap {{B}^{C}}$ on the first term of the above equation, we get
${{\text{A}}^{C}}\cap {{\text{B}}^{C}}\cap {{\text{B}}^{C}}$
Therefore, we get
${{\text{A}}^{C}}\cap ({{\text{B}}^{C}}\cap {{\text{B}}^{C}})$
Now, we know that the intersection of the same set complement is again the set complement, we get
${{\text{A}}^{C}}\cap {{\text{B}}^{C}}$
Again, we will apply de morgan’s law ${{(A\cup B)}^{C}}={{A}^{C}}\cap {{B}^{C}}$ in the above equation, we get
${{(A\cup B)}^{C}}=RHS$
Hence it is the correct option.
Option (3) states that $iii)\text{ }{{\text{A}}^{C}}\cap (A\cap B)\cap {{B}^{C}}=\phi $
So, we will first solve the left-hand side of the above equation, which is
${{\text{A}}^{C}}\cap (A\cap B)\cap {{B}^{C}}$
Now, we will first solve the first two terms of the above equation, by using the distributive property $(a\cap b)\cap c=(a\cap c)\cap (b\cap c)$ , we get
$\text{(}{{\text{A}}^{C}}\cap A)\cap ({{\text{A}}^{C}}\cap B)\cap {{B}^{C}}$
As we know, the intersection of a set with its complement is a null set, therefore we get
$\phi \cap ({{\text{A}}^{C}}\cap B)\cap {{B}^{C}}$
Now, again we will use the distributive property $(a\cap b)\cap c=(a\cap c)\cap (b\cap c)$on the last two terms , we get
$\phi \cap ({{\text{A}}^{C}}\cap {{B}^{C}})\cap ({{B}^{C}}\cap {{B}^{C}})$
As we know, the intersection of a set with its complement is a null set, therefore we get
$\phi \cap ({{\text{A}}^{C}}\cap {{B}^{C}})\cap \phi $
Thus, we will apply de morgan’s law ${{(A\cup B)}^{C}}={{A}^{C}}\cap {{B}^{C}}$ in the above equation, we get
$\phi \cap {{(\text{A}\cup B)}^{C}}\cap \phi $
As we know, the intersection of null set and the complement of the union of two sets is again the null set, therefore we get
$\phi =RHS$
Hence it is the correct option.
Option (4) states that $iv)\text{ (}{{\text{A}}^{C}}\cap B)\cap (A\cup B)=B$
So, we will first solve the left-hand side of the above equation, which is
$\text{ (}{{\text{A}}^{C}}\cap B)\cap (A\cup B)$
Now, we will know that $\text{ (}{{\text{A}}^{C}}\cap B)=B$ , therefore we get
$B\cap (A\cup B)$
Thus, we will apply the distributive property $(a\cup b)\cap c=(a\cap c)\cup (b\cap c)$ in the above equation, we get
$\text{ (B}\cap A\text{)}\cup \text{(B}\cap B\text{)}$
As we know that $\text{B}\cap B=B$ , therefore we get
$\text{ (A}\cap B\text{)}\cup B=B=RHS$
Hence it is the correct option.
Therefore, from the given option (i), (ii), (iii), and (iv) , all are the correct options.

Note: While solving this problem, do solve all the options properly because the question statement says that there is/ are a correct option which implies, we may have more than one correct option in this problem. Do mention all the properties properly to avoid an error.