1. Probability

Find an example in which:
P(AandB)>P(A)*P(B)

P(AandB)<P(A)*P(B)

2. Originally Posted by ulysses123
Find an example in which:
P(AandB)>P(A)*P(B)

P(AandB)<P(A)*P(B)
Hint: A and B are not independent events.

What have you tried? What thoughts have you had?

3. Hello, ulysses123!

Find an example in which: . $\begin{array}{cccc}[1] & P(A \cap B)&>&P(A)\!\cdot\!P(B) \\ \\[-3mm] [2] & P(A \cap B) &<& P(A)\!\cdot\!P(B)\end{array}$

We randomly select a number from: . $\{2,3,4,5\}$

We have: . $\begin{Bmatrix}P(\text{odd}) &=& \frac{2}{4} \\ \\[-3mm]
P(\text{even}) &=& \frac{2}{4} \\ \\[-3mm]
P(\text{prime}) &=& \frac{3}{4} \end{Bmatrix}$

$P(\text{odd}\cap\text{prime}) \:=\:\frac{2}{4} \:=\:\frac{1}{2}$

$P(\text{odd})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\cdot\frac{3}{4} \:=\:\frac{3}{8}$

Therefore: . $[1]\;P(\text{odd}\cap\text{prime}) \;>\;P(\text{odd})\!\cdot\!P(\text{prime})$

$P(\text{even}\cap\text{prime}) \:=\:\frac{1}{4}$

$P(\text{even})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\!\cdot\!\frac{3}{4} \:=\:\frac{3}{8}$

Therefore: . $[2]\;P(\text{even}\cap\text{prime}) \;<\;P(\text{even})\!\cdot\!P(\text{prime})$

4. 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.

5. Originally Posted by ulysses123
[snip]
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.
Is $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$ true when A and B are disjoint?

6. if they are disjoint the P(A and B)=0. Does this mean that the condition for independance can Never be fulfilled for disjoint events?

7. If two sets are disjoint (and they have some positive probability), then they are DEPENDENT.
This does make sense.
Since IF you know A has occured then you know B cannot.
Hence they do have a relationship.