Find an example in which:
P(AandB)>P(A)*P(B)
P(AandB)<P(A)*P(B)
Hello, ulysses123!
Find an example in which: .$\displaystyle \begin{array}{cccc}[1] & P(A \cap B)&>&P(A)\!\cdot\!P(B) \\ \\[-3mm] [2] & P(A \cap B) &<& P(A)\!\cdot\!P(B)\end{array}$
We randomly select a number from: .$\displaystyle \{2,3,4,5\}$
We have: .$\displaystyle \begin{Bmatrix}P(\text{odd}) &=& \frac{2}{4} \\ \\[-3mm]
P(\text{even}) &=& \frac{2}{4} \\ \\[-3mm]
P(\text{prime}) &=& \frac{3}{4} \end{Bmatrix}$
$\displaystyle P(\text{odd}\cap\text{prime}) \:=\:\frac{2}{4} \:=\:\frac{1}{2} $
$\displaystyle P(\text{odd})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\cdot\frac{3}{4} \:=\:\frac{3}{8}$
Therefore: .$\displaystyle [1]\;P(\text{odd}\cap\text{prime}) \;>\;P(\text{odd})\!\cdot\!P(\text{prime})$
$\displaystyle P(\text{even}\cap\text{prime}) \:=\:\frac{1}{4}$
$\displaystyle P(\text{even})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\!\cdot\!\frac{3}{4} \:=\:\frac{3}{8}$
Therefore: .$\displaystyle [2]\;P(\text{even}\cap\text{prime}) \;<\;P(\text{even})\!\cdot\!P(\text{prime}) $
what i have looked at is if the two events are dijoint, then the intersection is 0 and thus P(AandB)<P(A)*P(B)
But can someone explain to me why two disjoint events are actually NOT independant.Thsi concept has been in the back of my mind since the start of semester, and i havn't really figured it out yet.