If n(A)= 8 and n(A∩ B)=2, then n(A∩ B)' ∩ A ) is equal to

Question

If n(A)= 8 and n(A∩ B)=2, then n(A∩ B)' ∩ A ) is equal to (A) 2 (B) 4 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

Given, n(A)=8, n(A ∩ B)=2 n((A ∩ B) prime ∩ A)=n(A)-n(A ∩ B) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/fa6c440c4178b63dbb8e3830e956ef4a-.png" /> n(A ∩ B)=2 =8-2=6

Q. If $n (A) = 8$ and $n (A \cap B) = 2$ , then $n (A \cap B)^{'} \cap A)$ is equal to

$2$

$4$

$6$

$8$

$10$

Given, $n (A) = 8, n (A \cap B) = 2$
$n ((A \cap B)^{'} \cap A) = n (A) - n (A \cap B)$

$n (A \cap B) = 2$
$= 8 - 2 = 6$

Q. If n(A)=8 and n(A∩B)=2, then n(A∩B)′∩A) is equal to

2

4

6

8

10

Given, n(A)=8,n(A∩B)=2 n((A∩B)′∩A)=n(A)−n(A∩B) n(A∩B)=2 =8−2=6

Q. If $n (A) = 8$ and $n (A \cap B) = 2$ , then $n (A \cap B)^{'} \cap A)$ is equal to

$2$

$4$

$6$

$8$

$10$

Given, $n (A) = 8, n (A \cap B) = 2$
$n ((A \cap B)^{'} \cap A) = n (A) - n (A \cap B)$

$n (A \cap B) = 2$
$= 8 - 2 = 6$