If A and B are two events such that $P(A)=\frac{1}{2}$ and $P(B)=\frac{2}{3}$, then wjich of the following is incorrect ?

$P(A ∩ \overline{B}) ≥ \frac{1}{3}$

We have,

$P(A ∩ B) ≥ $ max $\begin{Bmatrix}P(A), P(B)\end{Bmatrix}=2/3$

and, $P(A ∩ B) ≤$ min $ \begin{Bmatrix}P(A), P(B)\end{Bmatrix}=1/2$

So option (a) is correct.

Now,

$ P(A  ∩ B) = P(A) + P(B) - P(A ∪ B)$

$⇒ P(A  ∩ B)≥ P(A) +P(B) -1 =\frac{1}{6}$

$∴ \frac{1}{6}≤ P(A  ∩ B)≤\frac{1}{2}$

So, option (c) is correct.

Now, $\frac{1}{6}≤ P(A  ∩ B)≤\frac{1}{2}$

$⇒ -\frac{1}{2} ≤ -(P ∩ B) ≤ -\frac{1}{6}$

$⇒ \frac{2}{3}-\frac{1}{2}≤ P(B) -P(A ∩ B) ≤ \frac{2}{3} -\frac{1}{6}⇒\frac{1}{6}≤ P(\overline{A}  ∩ B) ≤ \frac{1}{2}$

So, option (d) is correct.

Now, $P(A ∩ \overline{B})=  P(A) - p(A ∩  B)$

$⇒ P(A ∩ \overline{B} ) ≤ \frac{1}{2}-\frac{1}{6}=\frac{1}{3}$

so, option (d) is correct.