1. ## Spinning Wheel Probability

A spinning wheel is divided into 9 equal sections and each section is labeled with a number. The numbers from these sections are 1,2,3,4,5,6,7,8,9. Determine the probability that after n spins (n>1), the product of the n numbers gernerated will be divisible by 10.

Here's what I think:
In order to be divisible by 10, there has to be at least 2 spins. One spin has to be a 5 and the other spin has to be one of the even numbers (2,4,6,8). If I get these 2 combinations, it really doesn't matter what the other spins are. So I think the probability is (1/9) * (4/9) * ???

I'm not sure what the last ??? is. Any suggestions?

Thanks.

2. ## Re: Spinning Wheel Probability

Originally Posted by april19
A spinning wheel is divided into 9 equal sections and each section is labeled with a number. The numbers from these sections are 1,2,3,4,5,6,7,8,9. Determine the probability that after n spins (n>1), the product of the n numbers gernerated will be divisible by 10.

Here's what I think:
In order to be divisible by 10, there has to be at least 2 spins. One spin has to be a 5 and the other spin has to be one of the even numbers (2,4,6,8). If I get these 2 combinations, it really doesn't matter what the other spins are. So I think the probability is (1/9) * (4/9) * ???

I'm not sure what the last ??? is. Any suggestions?

Thanks.
I gather you are to solve generically for any integer n > 1. I suspect the problem statement says something about equal probability and independence.

This is a tricky problem, and I suspect there are several ways to solve it. Let's try one way, which may perhaps be the easiest.

If all n spins result in a 5, the product will be not evenly divisible by 10. The probability of this is obviously (1/9)[sup:3fxwpxzi]n[/sup:3fxwpxzi].

If none of the n spins results in a 5, the product will not be evenly divisible by 10. The probability of this is obviously (8/9)[sup:3fxwpxzi]n[/sup:3fxwpxzi].

If all n spins result in an even, the product will not be evenly divisible by 10. The probability of this is obviously (4/9)[sup:3fxwpxzi]n[/sup:3fxwpxzi].

If no spin results in an even, the product will not be evenly divisible by 10. The probability of this is obviously (5/9)[sup:3fxwpxzi]n[/sup:3fxwpxzi].

In every other case, you have at least one spin resulting in 5 and at least one resulting in an even, so the product will be evenly divisible by 10.

QUESTION TIME: Are the four cases described above mutually exclusive?

3. ## Re: Spinning Wheel Probability

I think the 4 cases are mutually exclusive because they can't happen at the same time.

4. ## Re: Spinning Wheel Probability

Originally Posted by april19
I think the 4 cases are mutually exclusive because they can't happen at the same time.
Ahh, but some of the cases CAN happen at the same time. The case of all n spins resulting in 5 entails that no spin results in an even. Similarly, the case

of all n spins resulting in an even entails that no spin results in a 5. So, the four cases are not mutually exclusive. I told you this one was tricky.

Can you see a way to calculate the probability that, out of n spins, a 5 resulted a times and that an even resulted at least (n - a) times,

where 1 <= a <= (n - 1).

Hint: Can you see a way to calculate the probability that, out of n spins, a 5 resulted a times and that an even resulted no times?

5. ## Re: Spinning Wheel Probability

Well, out of n spins, a 5 resulted in a times, I think, is (1/9)[sup:q1sk2w8c]a[/sup:q1sk2w8c].
An even number resulted in no times is (5/9)[sup:q1sk2w8c]n[/sup:q1sk2w8c].

I don't understand why "an even resulted no times" is an hint. Why wouldn't I want to find out an even resulted in (n-a) times? Wouldn't that be (4/9)[sup:q1sk2w8c]n-a[/sup:q1sk2w8c]?

Sorry for asking so many questions. Probability is not my strong point. It just doesn't come easy to me.
Thank you so much for helping.

6. ## Re: Spinning Wheel Probability

Originally Posted by april19
Well, out of n spins, a 5 resulted in a times, I think, is (1/9)[sup:s0pp2box]a[/sup:s0pp2box].
An even number resulted in no times is (5/9)[sup:s0pp2box]n[/sup:s0pp2box].

I don't understand why "an even resulted no times" is an hint. Why wouldn't I want to find out an even resulted in (n-a) times? Wouldn't that be (4/9)[sup:s0pp2box]n-a[/sup:s0pp2box]?

Sorry for asking so many questions. Probability is not my strong point. It just doesn't come easy to me.
Thank you so much for helping.
For many people, probability seems counter-intuitive at first. You just have to work at it a while, and your intuition develops.

Let's take this step by step OK?

Well, out of n spins, a 5 resulted in a times, I think, is (1/9)[sup:s0pp2box]a[/sup:s0pp2box].

That is not quite correct. Let's take an example. Say n is 3 and a is 2.
Event 1: you get a 5 on the first spin and a 5 on the second spin and a not-five on the third spin. Probability is (1/9) * (1/9) * (8/9).
Event 2: you get a 5 on the first spin and a not-five on the second spin and a five on the third spin.
Probability is (1/9) * (8/9) * (1/9) = (1/9) * (1/9) * (8/9).
Event 3: you get a not-five on the first spin and a 5 on the second spin and a 5 on the third spin.
Probability = (8/9) * (1/9) * (1/9) = (1/9) * (1/9) * (8/9).
Now events 1, 2, and 3 ARE mutually exclusive. So the probability of event 1 or event 2 or event 3 is just the sum of their individual probabilities.
So the probability = (1/9)(1/9)(8/9) + (1/9)(1/9)(8/9) + (1/9)(1/9)(8/9) = 3 * (1/9)[sup:s0pp2box]2[/sup:s0pp2box] * (8/9). You have three spins so you have three probabilities. Does that make sense?
The product of those probabilities is multiplied by the number of ways the joint event can happen, which is also 3 in this example.
Does that make sense?
The problem with examples is that you are asked to solve this for any n, not 3.
So the probability of exactly a 5s out of n spins is {n! / [a! * (n - a)!]} * (1/9)[sup:s0pp2box]a[/sup:s0pp2box] * (8/9)[sup:s0pp2box](n-a)[/sup:s0pp2box]. Do you understand why?
I suspect now that the ??? that you asked about in your first post is the factor {n! / [a! * (n - a)!]}.
Did you study combinations and permutations as the first step in learning probability theory?
{n! / [a! * (n - a)!]} tells you how many ways there are to get a successes out of n trials when you do not care about the order in which successes occur.

If you understand the above we can go on to the next step. This problem demands a lot of understanding so I do not want to go on until you are comfortable with the explanation for each step. I am also going to ask tkhunny to look at this thread. I studied probability theory forty years ago, and he can probably explain it better than I can (and he will not make any mistakes but I may).

7. ## Re: Spinning Wheel Probability

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

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.

I don't know how you get {n! / [a! * (n - a)!]} but after trying different n's and a's, it does give the number of different ways to hit the number after n trials. That's neat!

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?

Thank you for helping.

8. ## Re: Spinning Wheel Probability

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.

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.

9. ## Re: Spinning Wheel Probability

OK. I am stuck at your explanation.

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.

The original problem said the product of the n numbers generated will be divisible by 10. It didn't say what each spin can or cannot be. If I get a 5 on one spin and an even on another spin, why must the 3rd spin not a 5 or even?

10. ## Re: Spinning Wheel Probability

Originally Posted by april19
OK. I am stuck at your explanation.

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.

The original problem said the product of the n numbers generated will be divisible by 10. It didn't say what each spin can or cannot be. If I get a 5 on one spin and an even on another spin, why must the 3rd spin not a 5 or even?
I was going down a path that I was 100% confident would work. I was calculating the probability of exactly one 5 and exactly one even and saying that that probability did not take into account other sequences that gave a result evenly divisible by 10. So I was going to calculate the probability of exactly two 5s and one even and the probability of exactly one 5 and two evens and add the probabilities up.

P(one 5, one even, and one not) = [3 * (1/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] * [2 * (4/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] * [1 * (4/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] = 13.17%.
P(two 5s, one even, and zero nots) = [3 * (1/9)[sup:37mqr8w2]2[/sup:37mqr8w2]] * [1 * (4/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] = 1.65%.
P(one 5, two evens, and zero nots) = [1 * (1/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] * [3 * (4/9)[sup:37mqr8w2]2[/sup:37mqr8w2] = 6.58%.
P(result evenly divisible by 10 in three spins) = 21.40%.

P(one 5, one even, and one any) = [3 * (1/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] * [2 * (4/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] * [1 * (9/9)[sup:37mqr8w2]1[/sup:37mqr8w2]] = 29.63%.

We get different answers so one method must be wrong. I am now only 90% confident that my method is right, but I cannot explain why your method is wrong. That means I may be wrong. I am going to ask for more help.

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