Originally Posted by

**april19**
Thank you so much for being so patient with me and taking your time to explain it so well.

Let's reserve the thanks until I help you arrive at the correct answer. As I said, I studied basic probability theory probably before your parents were born, and I may screw up. We are working this problem together.

I am sorry to say that I have not taken any Combinations and Permutations classes before. I am only a High School student. This problem is an extra credit that can help me with my grade.

Usually, permutations and combinations are studied in the first one or two classes in basic probability theory. I do not know how anyone expects high school students to solve this problem without previously providing the formulas for permutation and combinations.

I don't know how you get {n! / [a! * (n - a)!]}

I have never proved it myself. I just memorized the formula long, long ago. I am sure the proof depends on mathematical induction, which you may not have learned yet.

but after trying different n's and a's, it does give the number of different ways to hit the number after n trials.

You have a great experimental attitude.

That's neat! Yes, it is.

OK. Got a question about your explanation.

If I spin and get one 5 and another spin gets an even number, then no matter what I get after n-2 spins, the product will be divisible by 10. So in your examples or explanation, why do I need to worry about a>1?

[color=#FF0000]GREAT QUESTION. Let's answer it in three parts.

First, even fairly simple problems in probability theory can frequently be solved in a variety of ways. Part of the trick is sort of seeing what the easiest way will be, but that's where intuition comes in. Until it is trained, it may often lead you astray so you need to think very carefully before relying on it.

Second, let's start with the simplest possible case, which is n = 2, AS YOU ALREADY SAW.

If you get a 5 on the first spin and an even on the second spin, the probability of that sequence = (1/9) * (4/9) = 4/81 = 4.94% approximately. That was the answer you came up with in your very first post. But that is not the only sequence that gives you what you want. If you get an even on the first spin and a 5 on the second spin that ALSO gives you what you want. The probability of that alternative sequence is (4/9) * (1/9) = 4.94%. Notice that you get one sequence or the other, but you cannot get both. They are MUTUALLY EXCLUSIVE. So the probability of getting one or the other of two mutually exclusive events is simply the sum of the probabilities of the two events. (You understand I trust that you cannot simply add probabilities if the events are not mutually exclusive.) So, the probability that you seek is (8/81) = 9.88% approximately.

Let's summarize what you learned from this. If you are going to calculate the probability by looking at the sequences that give you the desired result, you must find the probability for every sequence and, if you are sure that they are all mutually exclusive, you just add them all up. This way always works, but it is very inefficient except when n is small. Note by the way that when n = 2 and a = 1, {n! / [a! * (n - a)!]} = 2! / (1! * 1!) = 2, and 2 * (1/9)[sup:1th3k0e9]a[/sup:1th3k0e9] * (4/9)[sup:1th3k0e9](n - a)[/sup:1th3k0e9] gives the correct answer.

Let's try the next simplest case, which is n = 3.

AS YOU CORRECTLY OBSERVE, if you got a 5 on one spin and an even on another, the product would be evenly divisible by 10. That is, a = 1. I hope you see now why simply saying that getting a 5 on the first spin and an even on the second spin does not do the trick. How many ways can you get exactly one 5 out of three spins and exactly one even out of the two remaining spins. Well there are three ways to get exactly one 5, either on the first, second, or third spin. Having got exactly one five, there are two ways to get exactly one even from the remaining two spins, either on the first remaining spin or the second remaining spin. Having got one 5 and one even on two out of the three spins, you must not get a 5 or an even on a third spin if you are going to end up with exactly one five and one even. That means you must get a 1, 3, 7, or 9. They are mutually exclusive so their joint probability is (4/9). So, the probability of getting exactly one 5 and exactly one even = 3 * 2 * (1/9)[sup:1th3k0e9]1[/sup:1th3k0e9] * (4/9)[sup:1th3k0e9]1[/sup:1th3k0e9] * (4/9)[sup:1th3k0e9]1[/sup:1th3k0e9] = 13.17% approximately. Note that the sum of the exponents equals the number of spins.

BUT THAT IS NOT THE ONLY WAY OF GETTING A PRODUCT EVENLY DIVISIBLE BY 10.

You could also get exactly two fives and exactly one even or exactly one five and exactly two evens.

Considering just those sequences when you get exactly one 5 and exactly one even does not address the other sequences that also give you a product evenly divisible by 10.

Can you calculate the probabilties of those other ways?

So what is the probability that with 3 spins, the product is evenly divisible by 10.

BUT we have to solve this for any n > 1.

If you got n 5s, that product would not be divisible by 10 because there would be no even factor.

If you got no 5s, that product would not be evenly divisible by 10 because there is no factor of 5.

We need to find the probability of sequences that have a fives and b evens, where 1 <= a <= (n - 1) and 1 <= b <= (n - a).

Do you see why I said at the beginning that this problem is tricky? You must be in a very advanced high scool class.

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