I need help with homework!

[Phylum]

This is Yr.8 maths, so it should be easy for you!

I have to show a sample space (all possible outcomes) to a 'Balls in the Clowns Mouth' game (the non-dirty one)

For those who haven't played it, you have X amount of small plastic balls and place them in the mouth of a fake clown. The balls are then dispersed into Y compartments divided by walls with numeric values (1 to Y, often). At the end (when all balls are used up), the numeric values on the compartments are added up. Highest prizes are often awarded for the hardest to get.

[EXAMPLE] If there were 4 compartments, 1-10, and you had 8 balls, 8 and 80 would yield the best prizes.[/EXAMPLE]

In my case, I need to know the sample space for if there were 5 balls and 7 compartments, numbered 1 to 7.

I'd ask my maths teacher, but this is due Monday, and I've had it for a week.

Thanks!

Anyone else with hard homework want to make me feel less stupid?

[Strike Witch]

Just write "someone threw up in the machine" and hand it in.

[Wil]

What a quite bizarre concept. Are there different probabilites of a ball landing in any particular compartment? I'm guessing there is, since if there were an equal change, all 8 balls landing in Compartment 1 (scoring 8) would have exactly the same likelihood of occuring as all 8 balls landing in Compartment 2 (scoring 16), but the second apparently doesn't win you any prizes.

I'm also going to assume that each ball is independent: if Compartment Y is filled with X balls, then ball X+1 has exactly the same chance of landing in Compartment Y as any preceding ball.

I've been thinking a lot more about your problem (even diving back into my combinatorics of several years ago), but I can't think how best to proceed, and I also don't know how your work needs to be set out. If I were doing it casually, for myself, not intending to show my workings out to anyone, I'd do it something like this:

P(1), P(7) = 0.05
P(2), P(6) = 0.1
P(3), P(5) = 0.2
P(4) = 0.3

P(1×7) = 0.05⁷
P(1×6 + 2) = 0.05⁶ × 0.1
P(1×5 + 3) = 0.05⁵ × 0.2

And so on. I'm sure you can see what I'm doing. Trouble is it requires a hell of a lot of pencil mileage. Almost certainly this can be solved with combinations (C⁷₅) or permutations (P⁷₅) somehow, but I can't remember how to use them properly, and if you haven't been introduced to them, your teacher wouldn't expect you to use them.

You could shorten this section by simply stating the general formula, which would be (given the probabilities I've invented here):
P(1a + 2b + 3c + 4d + 5e + 6f + 7g) = 0.05^a+g × 0.1^b+f × 0.2^c+e × 0.3^d
Where a to e are the numbers of balls in each compartment

You can then add together the probabilities of all the outcomes that yeild the same score, for example, 10:
P(Score=10) = P(1×6 + 4) + P(1×5 + 2 + 3) + P(1×4 + 2×3)
= (0.05⁶ × 0.3) + (0.05⁵ × 0.1 × 0.2) + (0.05⁴ × 0.1³)
= 1.71875×10⁻⁸
= 1.72×10⁻⁸ (3 s.f.)

Which seems a stupidly small amount, but that's because the probabilities I've used are greatly exaggerated over those a real-life device of this nature would exhibit.

Of course, if there is equal probability of any ball landing in any compartment, then P(Score=Z) = 1/7 × the number of ball arrangements that score Z

I hope this helps at least a little, it's been thrown together haphazardly.

[Havoc]

What a quite bizarre concept. Are there different probabilites of a ball landing in any particular compartment? I'm guessing there is, since if there were an equal change, all 8 balls landing in Compartment 1 (scoring 8) would have exactly the same likelihood of occuring as all 8 balls landing in Compartment 2 (scoring 16), but the second apparently doesn't win you any prizes.

I'm also going to assume that each ball is independent: if Compartment Y is filled with X balls, then ball X+1 has exactly the same chance of landing in Compartment Y as any preceding ball.

I've been thinking a lot more about your problem (even diving back into my combinatorics of several years ago), but I can't think how best to proceed, and I also don't know how your work needs to be set out. If I were doing it casually, for myself, not intending to show my workings out to anyone, I'd do it something like this:

P(1), P(7) = 0.05
P(2), P(6) = 0.1
P(3), P(5) = 0.2
P(4) = 0.3

P(1×7) = 0.05⁷
P(1×6 + 2) = 0.05⁶ × 0.1
P(1×5 + 3) = 0.05⁵ × 0.2

And so on. I'm sure you can see what I'm doing. Trouble is it requires a hell of a lot of pencil mileage. Almost certainly this can be solved with combinations (C⁷₅) or permutations (P⁷₅) somehow, but I can't remember how to use them properly, and if you haven't been introduced to them, your teacher wouldn't expect you to use them.

You could shorten this section by simply stating the general formula, which would be (given the probabilities I've invented here):
P(1a + 2b + 3c + 4d + 5e + 6f + 7g) = 0.05^a+g × 0.1^b+f × 0.2^c+e × 0.3^d
Where a to e are the numbers of balls in each compartment

You can then add together the probabilities of all the outcomes that yeild the same score, for example, 10:
P(Score=10) = P(1×6 + 4) + P(1×5 + 2 + 3) + P(1×4 + 2×3)
= (0.05⁶ × 0.3) + (0.05⁵ × 0.1 × 0.2) + (0.05⁴ × 0.1³)
= 1.71875×10⁻⁸
= 1.72×10⁻⁸ (3 s.f.)

Which seems a stupidly small amount, but that's because the probabilities I've used are greatly exaggerated over those a real-life device of this nature would exhibit.

Of course, if there is equal probability of any ball landing in any compartment, then P(Score=Z) = 1/7 × the number of ball arrangements that score Z

I hope this helps at least a little, it's been thrown together haphazardly.

[MA]

FUCK THIS THREAD YOU MADE ME FEEL INADEQUATE AGAIN

NO MR HETBURN I WON'T LEAVE FUCK YOU

[Mac Sirloin]

Wait, so how does this involve taking apart blenders?

[OANST]

Wait, so how does this involve taking apart blenders?

You have to carry the three.

[MA]

whoa, no one mentioned there being more than one bender.

[Daxter King]

What the fuck kind of year 8 math is this?

[Mac Sirloin]

You have to carry the three.

Three? Three what? Size Three Torux head? Carry it where? I need it to take the blender apart!

[Josh]

Phylum, if you are 14, why are you doing Year 8 math?

[Mac Sirloin]

I was 14 In Grade 8.

Oh wait

No

I was 13.

Maybe it was his birthday

[enchilado]

1) Divide the number of compartments by how many balls you have.
2) Multiply by your postcount.
3) Subtract 0.
4) Stain your homework with coffee rings.
5) Put it through one of the blenders.
5) Feed it to the dog.
6) Hand in a "Say 'I do' Mr Rudd" pamphlet instead.



Hope this helps

[Nate]

It occurs to me that the slots on the far left and far right have a lower chance of receiving a ball than the other ones, as they're only passed over once once on each back-and-forth path.

If you ignore that, the number of possible outcomes is 7⁵.

[Phylum]

Nate, within he numbers 5-35?

same likelihood of occuring as all 8 balls landing in Compartment 2 (scoring 16), but the second apparently doesn't win you any prizes.

This was due to my example, will be more clearly labeled shortly.

In my case, I need to know the sample space for if there were 5 balls and 7 compartments, numbered 1 to 7.

Balls, 5. Compartements, 7.

[Wil]

It occurs to me that the slots on the far left and far right have a lower chance of receiving a ball than the other ones, as they're only passed over once once on each back-and-forth path.

If you ignore that, the number of possible outcomes is 7⁵.

Back and forth path? Are you familiar with the shape and structure of this clown contraption in a way I'm not?

Balls, 5. Compartements, 7.

Yeah, I just did a whole great post with 5 balls and 7 compartments.

[Nate]

Nate, within he numbers 5-35?

What?

[Phylum]

It surprises me to have 16,807 different ways to make the numbers 5-35 using 5 numbers from 1-7. Very poorly articulated by me, this time and last...

Yeah, I just did a whole great post with 5 balls and 7 compartments.

Sorry, hadn't bothered to read it when I posted. I did soon after and thought man, I'm such a dick. I read the first bit and thought: I'll address that now, else I might forget!

[Wil]

It's entirely true.

I just looked up the term "sample space" (admittedly only on Wikipedia), and all it is, is the set of all possible outcomes. In the case of your homework, the sample space is:

{5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35}

[Nate]

It surprises me to have 16,807 different ways to make the numbers 5-35 using 5 numbers from 1-7. Very poorly articulated by me, this time and last...



Sorry, hadn't bothered to read it when I posted. I did soon after and thought man, I'm such a dick. I read the first bit and thought: I'll address that now, else I might forget!

There are 16,807 different ways if you consider that order is important; ie 1-2-3-4-5= 15 is different to 5-4-3-2-1 and 1-3-5-2-4, etc. If order is unimportant, then the number is much smaller but I don't know how to calculate that.

[Phylum]

The number, not the order, is what's being considered here.

Thanks, I'm gonna try to attempt it now. I probably misinterpreted the question, due to the complexity of the answers.

[Nate]

In that case, Max's last post should be the answer.

[enchilado]

No. Max's last post was the number of combinations if the order was important.
Say, you've got 5 balls number 1, 2, 3, 4, 5. And Max's answer is if 2, 1, 3, 4, 5 is different to that.

Phylum wants the figure that says they're the same.

[Nate]

I was talking about this post, in which Max gives the list of possible final scores for five balls:

It's entirely true.

I just looked up the term "sample space" (admittedly only on Wikipedia), and all it is, is the set of all possible outcomes. In the case of your homework, the sample space is:

{5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35}

[Wil]

No. Max's last post was the number of combinations if the order was important.
Say, you've got 5 balls number 1, 2, 3, 4, 5. And Max's answer is if 2, 1, 3, 4, 5 is different to that.

No, my first post considers that the balls as independent of each other, thus their order doesn't matter. The probabilities of getting each final score are derived by adding up the probabilities of each possible end scenario, i.e. how many balls are in each compartment, not which balls are in which compartment.

[MA]

why the fuck are we having a debate over year 8 maths homework?

i cant stop laughing.

"I like purple."
"Well, purple is supposed to signify sophistication."
"I didn't say I am defined as purple, I just said I like it."
"Liking it is the reason it defines you."
"Read my above post. I like it. I am not defined as it."
"Hope this helps. "
and so on.

[Wil]

The real question is, why is this one of the more interesting and cohesive discussions OWF has had in a while. Therein lies some curious ideas.

[Nate]

Because it's intellectually stimulating. How many recent threads can say that?

[Havoc]

At least two.

[Nate]

You have a strange definition of the word 'recent'.