proability

[tushar7]

if i pick a positve integer less then 1000 , what is the probability that if i summed it with a another postive integer less 1000 , the result will be divisible by 6?

[Avik Roy]

Is this problem equivalent to "What is the probability that sum of two integers, both less that 1000, is divisible by 6?"

[tushar7]

i guess so

[Zzzz]

We can divide the numbers into 6 disjoint sets $A_0,A_1,A_2,A_3,A_4,A_5$.
Where $A_i=\{ x:0<x<1000,x\equiv i(mod\ 6)\}$
So, $|A_1|=|A_2|=|A_3|=167$ and $|A_4|=|A_5|=|A_0|=166$
Its clear that if the first number comes from $A_0$, second number must come from $A_0$.
Probability=$\frac{166}{999}\cdot \frac{165}{998}$
First number from $A_1$, second number from $A_5$, probability=$\frac{167}{999}\cdot \frac{166}{998}$
First number from $A_2$, second number from $A_4$, probability=$\frac{167}{999}\cdot \frac{166}{998}$
First number from $A_3$, second number from $A_3$, probability=$\frac{167}{999}\cdot \frac{166}{998}$
First number from $A_4$, second number from $A_2$, probability=$\frac{166}{999}\cdot \frac{167}{998}$
First number from $A_5$, second number from $A_1$, probability=$\frac{166}{999}\cdot \frac{167}{998}$

Answer=$\frac{166}{999}\cdot \frac{165}{998}+5\cdot \frac{167}{999}\cdot \frac{166}{998}$

[Avik Roy]

I think repetitive cases should be included

[Zzzz]

I think repetitive cases should be included

বুঝি নাই

[Avik Roy]

It's not necessary to assume that the two "0 modulo 6" numbers are different. So that probability should be $\left ( \frac{166}{999} \right )^{2}$

[Zzzz]

if i pick a positve integer less then 1000 , what is the probability that if i summed it with a another postive integer less 1000 , the result will be divisible by 6?

He said, another positive integer