Probability that the product of 5 numbers will be divisible by 9.

[pony1983]

1. The problem statement, all variables and given/known data
If 5 numbers are randomly generated (excluding zero) what are the odds that the product of the 5 numbers will be divisible by 9?

I realize that if at least one number is divisible by 9 then the product will also be divisible by 9 or if there's at least 2 numbers divisible by 3(that aren't divisible by 9) the product will be divisible by 9.



2. Relevant equations
I don't know the relevant equations.


3. The attempt at a solution
My best attempt at a solution is 1-(7/9)^4 - I feel like this would be the odds of getting at least 2 numbers that are divisible by 3 that aren't divisible by 9 and 1-(8/9)^5 I feel are the odds of getting at least 1 number divisible by 9. I don't know how to combine the 2 though.

P.S. This is not a homework problem but I felt this would be the proper forum as I believe it is a "homework style" question. Thanks for your help.

[Stephen Tashi]

You need to define how you intend to randomly generate the numbers. I assume you mean non-negative integers. Any limit on their size? Generating a random non-negative integer that can be arbitrarily large is an interesting task.

[pony1983]

Oh yeah sorry I meant non-negative integers, and no there's no limit on the size of the integer.As for how it is randomly generated , I don't know , just assume that it is random.

[Stephen Tashi]

probability of ( A or B) = probability of A + probabiity of B - probability of (A and B)

A = you get at least two numbers divisible by 3 , each of which is a number that is not divisible by 9

B = you get at least one number divisible by 9

So you need to work on computing the probability of (A and B) both happening. You might have to break the way it happens into several cases.

[pony1983]

Ok based on that info here is my work:
1-(7/9)^4= .3659 (probability of getting at least 2 numbers divisible by 3 but not by 9)
1-(8/9)^5= .5549 (probability of getting at least 1 number divisible by 9)
.3659 + .5549= .9208
.3659 x .5549= .2030 (probability of getting at least 2 numbers divisible by 3 but not by 9; and at least 1 number divisible by 9)

.9208 - .2030 = 71.78%

Have I done this problem correctly?

[Stephen Tashi]

probability of (A and B) = ( probability of (A given B) ) (probability of B)

If A and B are "independent events", probability of (A given B) = probability of A so you can multiply the probabilities of A and B.

In your problem, A and B are not independent events. If you get at least one number divisible by 9, then then you only have 4 or fewer chances to also get at least two numbers divisible by 3 and not by 9.

[pony1983]

I have to admit I'm completely lost on what to do now.I guess I'll sleep on it and see if i can figure it out.

[I like Serena]

As proposed the calculation will become rather complex.

I suggest an alternative approach:

P(product divisible by 9) = 1 - P(product not divisible by 9)

P(product not divisible by 9) = P(none of the 5 numbers is divisible by 3) + P(exactly 1 of the 5 numbers is divisible by 3, but not by 9)

Can you calculate those?