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 ?
Answer & explanation
Correct answer: option 2
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.