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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 )?

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

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  • $\begingroup$ thank you, your ans improved my way of thinking...! $\endgroup$ – raghu Apr 17 '13 at 6:29
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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.

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    $\begingroup$ What do you mean by independence if not statistical independence? $\endgroup$ – Did Apr 4 '13 at 11:57
  • $\begingroup$ @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. $\endgroup$ – A.D Apr 4 '13 at 12:18
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    $\begingroup$ Zero-covariance is absence of correlation, not independence. $\endgroup$ – Did Apr 4 '13 at 13:14

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