Three numbers, x, y, z ∈ N are chosen at random such that 1 ≤ x, y, z ≤ 10 ?

Probability Level 2

Hi, I am trying to understand the synopsis for a detailed solution, but I can’t.

Here is the body of the solution itself:

First Qeust: Did I understand the wording correctly?

Second Qeust: What are the two numbers? I can make the assumption that the first number is 9, but the second I can not assume. Since I am looking for the second number which after division gives 3. P.S. formula of incidence of the second number: $\frac{xyz}{3^{n}} = 3^{1}$

Third question: We can choose only one of these for our variables. Which variable can I take from the two current variables?

Fourth question: The other two variables must come from the set we considered in case 1. We are talking about two variables that, when divided, give a result of 3 with degree 1 or, more simply, give a result of 3?

The answer is 0.358.

1 solution

Chris Lewis
Jan 10, 2020

It would be a bit better to post questions like this on the discussion part of this site, since you're after an explanation of a given solution.

However, the idea of the solution you've found is counting the number of times $3$ occurs in the prime factorisation of the product of three numbers.

The solution categorises each number $1$ to $10$ in terms of the highest power of $3$ that divides into it.

The numbers in $A=\{1,2,4,5,7,8,10\}$ are not divisible by $3$ (so the highest power is $3^0=1$ ).

The numbers in $B=\{3,6\}$ are divisible by $3$ , but not by $9$ (so the highest power is $3^1=3$ ).

Finally, the number in $C=\{9\}$ is, of course, divisible by $9=3^2$ .

If we want to find products $xyz$ that are not divisible by $9$ , we have two choices: pick all three of $x,y,z$ from set $A$ , or pick one from set $B$ and two from set $A$ . Any more than two from set $B$ and the product will be divisible by $9$ .

The rest of the problem is then just counting these two cases.

By the way, a nice extension problem might be to change the initial set of numbers $x,y,z$ are chosen from; instead of $1$ to $10$ , what happens if they're chosen from $1$ to $100$ ? What do you expect the probability to be as the size of this set increases? Can you prove it?

(43/120)^2 ?

Saint Hentai - 1 year, 4 months ago

So if the set were $1$ to $1000000$ , the probability would be $\left(\frac{43}{120}\right)^6$ ? This suggests the probability tends to zero, but does that make sense?

You might want to think about the approximate sizes of the sets $A,B,C$ for a general (large) $n$ , where $n$ is the upper bound of choices for $x,y,z$ . For example, the size of $A$ is about $\frac{2n}{3}$ , and this approximation gets (in general) better and better for larger and larger $n$ .

Chris Lewis - 1 year, 4 months ago

Three numbers, x, y, z ∈ N are chosen at random such that 1 ≤ x, y, z ≤ 10 ?

The answer is 0.358.

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