If $P(A)=\frac{6}{11},P(B)=\frac{5}{11}$ and $P(A ∪ B)=\frac{7}{11},$ then

A. $P(A ∩ B)=\frac{4}{11}$

B. $P(A | B)=\frac{4}{5}$

C. $P(B | A)=\frac{2}{5}$

D. A and B are independent evennts

E. P(neither A nor B) $=\frac{4}{11}$

A, B and E only

The correct answer is Option (2) → A, B and E only

$P(A)=\frac{6}{11},P(B)=\frac{5}{11}$, $P(A ∪ B)=\frac{7}{11}$

so $P(A∩B)=P(A)+P(B)-P(A ∪ B)=\frac{4}{11}$

$P(A).P(B)≠P(A∩B)$ ⇒ A, B not independent

$P(A|B)=P(A)=\frac{6}{11},P(B|A)=P(B)=\frac{5}{11}$

P(neither A nor B) = $P(\overline{A ∪ B})=\frac{4}{11}$

A, B and E only are true