Probability that at least $1$ out of $10$ random numbers between $1 \ldots 1000$ is not divisible by $7$

$10$ different numbers are randomly chosen from the numbers $1, 2, \ldots, 1000$. What is the probability that at least one of the chosen numbers is not divisble by $7$?

If there are $x$ numbers to choose from, there are $x/7$ numbers that are divisible by $7$. Probability of a number being divisble by $7$ is always $(x/7)/x = 1/7$.

• $A$ - at least one number is not divisble by $7$
• $\bar{A}$ - all numbers are divisible by $7$

These numbers were chosen independently so is $P(\bar{A}) = (\frac{1}{7})^{10}$ and so $P(A)=1-P(\bar{A})=1-(\frac{1}{7})^{10}$

Is this correct?

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1 Answer

In the set $\{1, 2, \ldots, n\}$, there are precisely $\lfloor \frac{n}{7} \rfloor$ numbers that are divisible by $7$, not $\frac{n}{7}$.

Under the assumption that the same number can be chosen more than once, the rest seems correct.

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No, 10 different numbers are chosen. But since the probability is always $1/7$ no matter how many numbers I can choose from, the answer is the same, no? – mak Apr 5 '13 at 18:27
@mak: If I choose a random number between $2$ and $5$, then the probability that it is divisible by $7$ is also not $1/7$. Similarly, since $1000$ is not a multiple of $7$, here the probability is not exactly $1/7$. – TMM Apr 5 '13 at 18:32
@TMM: I choose the first number: 1000 numbers to choose from, probability of that number being divisble by 7 is $(1000/7)/1000=1/7$. I choose the second number: 999 numbers left to choose from, probability of that number being divisble by 7 is also $(999/7)/999=1/7$, and so on. My answer is still correct, no? (I know about rounding down the fraction) – mak Apr 5 '13 at 18:54
@mak It's not quite correct. Consider the set $\{0, 1, \ldots, 10\}$. Surely, the probability of choosing a number divisible by $7$ from that set is not $1/7$, right? – Sam Apr 5 '13 at 18:59
@mak Sorry, I meant the set $\{1, 2, \ldots, 10\}$. The point I'm trying to make is that the probability of choosing a number divisible by $7$ from that set is not $1/7$ because only one of the numbers (not $10/7$ of them) are divisible by $7$. Similarly, the probability of choosing a number divisible by $7$ from the set $\{1, 2, \ldots, 1000\}$ is not exactly $1/7$; it is $142/1000$. – Sam Apr 5 '13 at 19:28