# Independence in Probability

Consider $A$ set $A=\{1,2,3,4,5,6,7,8,9,10\}$, now event $E_1$ is getting a Number divisible by $2$, i.e., $P(E_1)=1/2.$ event $E_2$ is getting a Number Divisible by $3$, i.e., $P(E_2) = 3/10.$ Now $E_3$ is getting a number divisible by $6$, i.e., $P(E_3)=1/10;$

clearly $E_1,E_2$ are Independent Events & $E_3=E_1 \cap E_2$ we know that $P(E_3)=P(E_1)\cdot P(E_2)$ from probability theory...

if we apply in our problem

$$\frac1{10} = \frac3{20}$$

we are getting wrong why....what is the wrong in our consideration...(i mean my consideration..:D )?

• Are you certain that $E_1$ and $E_2$ are independent events? – Stefan Hansen Apr 4 '13 at 11:20

$E_1$ and $E_2$ are not independent, because there are more odd multiples of 3 than even ones in your set. Hence, if you know that a number randomly chosen is even, it is less likely to be a multiple of 3.
If you instead consider the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$, you find that half of the multiples of 3 are even and half are odd – so in this case, the two events would be independent.
Here $P(E_1 \cap E_2)=\frac{1}{10}\neq\frac{3}{20}=P(E_1)P(E_2)$. So, the events are not statistically independent. And you can apply $P(E_1 \cap E_2)=P(E_1)P(E_2)$ if they are statistically independent.
• @Did: I think about independence of variable and independence of event. Like $Y=X^2$ for $\{-1,0,1\}$. $Cov(X,Y)=0$ but the variables are not independent. But here it is not applicable. – A.D Apr 4 '13 at 12:18