Answer
(a) S={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(d,n,d,d),(d,n,d,n),(d,n,n,d),(d,n,n,n),(n,d,d,d),(n,d,d,n),(n,d,n,d),(n,d,n,n),(n,n,d,d),(n,n,d,n),(n,n,n,d),(n,n,n,n)}
(b) No, because $A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \neq \varphi$
(c) $A_{1}$ ={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(d,n,d,d),(d,n,d,n),(d,n,n,d),(d,n,n,n)}
(d) $A_{1}^{'}$={(n,d,d,d),(n,d,d,n),(n,d,n,d),(n,d,n,n),(n,n,d,d),(n,n,d,n),(n,n,n,d),(n,n,n,n)}
(e) $A_{1} \cap A_{2} \cap A_{3} \cap A_{4} $={(d,d,d,d)}
(f) $\left(A_{1} \cap A_{2}\right) \cup\left(A_{3} \cap A_{4}\right)=\{(d,d,d,d), (d,n,d,d), (d,d,d,n),(n,d,d,d),(d,d,n,d), \text {(n,n,d,d),(d,d,n,n)}\}$
Work Step by Step
Let "d" denote a distorted bit, Let "n" denote a not distorted bit.
(a) S={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(d,n,d,d),(d,n,d,n),(d,n,n,d),(d,n,n,n),(n,d,d,d),(n,d,d,n),(n,d,n,d),(n,d,n,n),(n,n,d,d),(n,n,d,n),(n,n,n,d),(n,n,n,n)}
Note that,
$A_{1}$={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(d,n,d,d),(d,n,d,n),(d,n,n,d),(d,n,n,n)}
$A_{2}$={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(n,d,d,d),(n,d,d,n),(n,d,n,d),(n,d,n,n)}
$A_{3}$={(d,d,d,d),(d,d,d,n),(d,n,d,d),(d,n,d,n),(n,d,d,d),(n,d,d,n),(n,n,d,d),(n,n,d,n)}
$A_{4}$={(d,d,d,d),(d,d,n,d),(d,n,d,d),(d,n,n,d),(n,d,d,d),(n,d,n,d),(n,n,d,d),(n,n,n,d)}
(b) No, because $A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \neq \varphi$
(c) $A_{1}$ ={(d,d,d,d),(d,d,d,n),(d,d,n,d),(d,d,n,n),(d,n,d,d),(d,n,d,n),(d,n,n,d),(d,n,n,n)}
(d) $A_{1}^{'}$={(n,d,d,d),(n,d,d,n),(n,d,n,d),(n,d,n,n),(n,n,d,d),(n,n,d,n),(n,n,n,d),(n,n,n,n)}
(e) $A_{1} \cap A_{2} \cap A_{3} \cap A_{4} $={(d,d,d,d)}
(f) $\left(A_{1} \cap A_{2}\right) \cup\left(A_{3} \cap A_{4}\right)=\{(d,d,d,d), (d,n,d,d), (d,d,d,n),(n,d,d,d),(d,d,n,d), \text {(n,n,d,d),(d,d,n,n)}\}$
