For two events A & B, which of the following is/are correct

(A \cup B) \cap B = B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(A \cup B)^{C} \cap B^{C} = (A \cup B)^{C}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A^{C} \cap (A \cap B) \cap B^{C} = ϕ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(A^{C} \cap B) \cap (A \cup B) = B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $(A \cup B) \cap B = B$
B $(A \cup B)^{C} \cap B^{C} = (A \cup B)^{C}$
C $A^{C} \cap (A \cap B) \cap B^{C} = ϕ$
Given $A, B$ are two events.
1) $(A \cup B) \cap B = (A \cup B) + B - ((A \cup B) \cup B)$ $\Rightarrow (A \cup B) \cap B = (A \cup B) + B - (A \cup B) = B$
$\Rightarrow (A \cup B) \cap B) = B$
2) $(A \cup B)^{c} \cap B^{c} = ((A^{c} \cap B) \cup B^{c}) = (A^{c} \cap B^{c})$
$\Rightarrow (A \cup B)^{c} \cap B^{c} = (A^{c} \cap B^{c}) = (A \cup B)^{c}$
3) $A^{c} \cap (A \cap B) = ϕ$ (since $A^{c}$ indicates excluding $A$ everything in the set, whereas $A \cap B$ indicates common between $A$ and $B$ . so intersection of both is a null set)
$A^{c} \cap (A \cap B) \cap B^{c} = ϕ \cap B^{c} = ϕ$
4) $(A^{c} \cap B) \cap (A \cup B) = (B - (A \cap B)) \cap (A \cup B) = (B - (A \cap B)$

Suggest Corrections