HELPPPPPPP

[Centurion1]

Need help with some stats work. I cannot stand the class. I got a very difficult professor who on top of it is from the Philippines and speaks horrendous english. All the tutors were taken and I am slowly becoming enraged.

I was wondering if to start off anyone could help me work through this problem.


Ely is on a 1000-calorie/day diet which she has been strictly following for about 2 weeks now. Thelma, her college buddy, is in town, though, and she has brought with her Ely’s favorite chocolate from their college town. Ely has a real weakness for chocolates and she will eat any type as long as it does not have any liquor in it. If Thelma brought a box of thirty indistinguishable chocolate nuggets (each one having 50 calories) and there are equal amounts of the 6 flavors—orange, hazelnut, rum, strawberry, almond and praline—what is the probability that she will consume 80% of her calorie allowance for the day on Thelma’s chocolates before she gets a rum-flavored chocolate?

[Crazed Rabbit]

So, we need to find the probability that she eats 16 other flavored chocolates before eating a rum flavored one.

There should be some sort of equation for this, but it's been a while since I've done this. It's straightforward in Excel though.

So the chance of her eating an "other" flavored chocolate is 25/30 for her first one ~ 83.3%. The chance of her next chocolate also being an other flavor gets slightly trickier. After she eats the first one, there's 5 rum and 24 other left. So by itself choosing an other is 24/29 or ~82.8%. However, the two events are connected. Her chance of choosing two others in a row is 88.3% * 82.8%, or ~69%. I think of the 69% as the cumulative chance she eat x number (2 in this case) other chocolates in a row. And so it goes, until she's eaten 16 other chocolates, and then we use the chance of her eating a rum one multiplied by the current cumulative chance of eating 16 others in a row.

I think, anyways.

CR

[Montmorency]

Eh, I get 1.4% with combinations. The chocolates are randomly chosen, correct? And it's just 80%, not at least 80%?

But I'm not very good at this sort of thing...

[Crazed Rabbit]

I believe the thing is that she eats 80% of her calories - 800, or 16 chocolates - before she tastes a rum one.

CR

[a completely inoffensive name]

From my poor understanding, I believe CR is right. In a more formal way the answer should be the product of ((n-5)/n) where n=number of chocolates from 30 to 14.

[Fragony]

Make one of these https://upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Normal_approximation_to_binomial.png/275px-Normal_approximation_to_binomial.png and a zero equation, set axis on 0.8 et presto

[Husar]

Well, since she won't put any of them back into the box after eating them, it should be a Hypergeometric distribution (http://en.wikipedia.org/wiki/Hypergeometric_distribution).

You can replace it with a Binomial distribution if n/N is > 0.05 but I don't think it's the case here.

So you could set:

N = 30
m = 5
n = 16

And set k = 0, which would get you the chance that if she picks 16 chocolates from 30 (5 of which have rum), there will be 0 with rum among the 16 she picked.

It doesn't say anything about whether she will eat one with rum directly afterwards but neither does the task description IMO.

The important part here being that she doesn't put any of the chocolates back and the Hypergeometric distribution is the only discrete distribution we learned that works when you don't put back what you picked out of N.

Edit: Thinking about it again, CR may be right that it implies the 17th will be a rum chocolate nugget, in that case you will have to do it manually I guess. There may be a chance that you could use the geometric distribution as well because N is 30 or so but that seems a bit flaky/inaccurate because it assumes she puts them back after eating them.

Or you could calculate the chance that she eats 16 non-rum nuggets and then calculate the chance that when she picks 1 from the remaining 14 nuggets, it's a rum nugget, then combine the two. Which is how CR did it except you'd speed up the calculation for the first 16 a bit by using the Hypergeometric distribution I guess.

[Centurion1]

wait should i multiply it out or should i use the hypergeometric process. also with the hypergeometric..... confused that k=0 because then it means that you get 5/0 which is undefined.

the point of the question... it is not a trick question in any way its number one for gods sakes and this professor is not going to give us trick questions. All she wants to know is th probability that you will not choose a rum flavored candy by the time she eats 80% of her calories which is 16 candies worth.

so just forget the 17th its irrelevant

[Crazed Rabbit]

My earlier method or the hypergeometric method give the same result. The equation I used was this;
https://upload.wikimedia.org/wikipedia/en/math/f/9/2/f928dfa53692944d1afe4204a2837368.png

Excel has a function for binomial coefficients - "Combin(top number, bottom number)"

It works thusly;
https://img412.imageshack.us/img412/7817/za006051122.gif

Yay factorials.

For the hypergeometric, k is the number of successes, n is the number of draws, N is the initial pool, and m is the initial number of what you're trying to pick out.

Remember that for this we're looking at picking 16 other chocolates in a row. So we want to consider picking the other chocolates as a success. Also, even with the hypergeometric equation it's a two part problem - the equation can't give you the answer for picking 16 others in a row and then picking a rum flavored chocolate.

CR

[Centurion1]

right we havent been using excel so its all sort of blurry.

wheni try to do the factorial i get massive unworkable numbers.


and so your saying that k=16; not zero?


the seventeenth candy is absolutely irrelevant. all they want to know is what the probability of NOT selecting a rum candy by the sixteenth

[Centurion1]

Never mind i figured it out. I reached the 1.4% using both methods, i.e. the hypergeometric and the multiplying out manually.

I appreciate the help. Unfortunately I have more to go and other problems i have found issue with.....

Does anyone have any idea on this one?


2. Photocards-R’-Us is a photo card printer. For this year, they have found that 10% of the cards they make are of inferior quality (blurred picture, faded colors, etc.) To attract customers to have their holiday cards printed with them, they are offering a 50% discount to customers who order a box of 50 cards if 10 of the cards in the box are of inferior quality. Manny is going to order a box of 50 Christmas photo cards at Photocards-R’-Us. What is the probability that he will receive a 50% discount on his holiday cards? [10 pts.]


Personally, I thought that it would be like Bayes' but the wording confuses me as well as the application required.

[Montmorency]

Grah!

Ha.

Complicated math speak here. But used simple combinations to get answer way up there.

2. Rephrase: Probability of getting at least 10 defective cards.

Binomial expansion: Summation for 10-50 cards defective. But simpler summation 0-9 cards defective. Call defective a failure.

p = probability success = .9
q = probability failure = .1
n = # times = 50
x = # successes
n - x = # failures


..............(x=50)
(p + q)^50 = Σ nCxpxq(n-x)
................(x=41)


And subtract from 1 to get probability you want. Can't think of simpler way. :shrug:

[Centurion1]

hrmmmm wat?

[Centurion1]

wait is this poisson because ive never seen it in that context

[Montmorency]

You've been poisoned?

[Centurion1]

poisson is a formula.

if you could elaborate i would really appreciate it in all seriousness.

[Montmorency]

What is the probability of getting a 50% discount, when a discount is given for 10/50 cards (at least) defective. 10% probability per card of defectiveness.

Sum probabilities of getting 1, 2, 3, 4, 5, 6, 7, 8, or 9 defective cards and subtract from 1. Or sum probabilities of 10, 11, 12...or 50 defective cards.

Use binomial expansion?

General binomial expansion: nC0p0qn + nC1p1qn-1 ... + nCn-1pn-1q1 + nCnpnq0

[Centurion1]

havent learned binomial expansion as strange as it may sound. what does the c mean?

[Montmorency]

choose.

You probably saw it as (pq). But that notation is ugly.

[Centurion1]

i unfortunately do not understand this at all.

[Centurion1]

is it a zscore problem?

[Husar]

My earlier method or the hypergeometric method give the same result. The equation I used was this;
https://upload.wikimedia.org/wikipedia/en/math/f/9/2/f928dfa53692944d1afe4204a2837368.png

My dear CR, that is the formula used with the Hypergeometric method. ~;)
The hypergeometric method is just a shorter way of doing what you did, I didn't want to sound like what you posted at first was wrong.


confused that k=0 because then it means that you get 5/0 which is undefined.

Then you're doing the binomial coefficients wrong, there shouldn't be a 5/0.


All she wants to know is th probability that you will not choose a rum flavored candy by the time she eats 80% of her calories which is 16 candies worth.

so just forget the 17th its irrelevant

Then just use the hypergeometric method. ~;)


right we havent been using excel so its all sort of blurry.

wheni try to do the factorial i get massive unworkable numbers.

On your calculator, to calculate a binomial coefficient (http://en.wikipedia.org/wiki/Binomial_coefficient) (one number above another in parentheses) you can type the upper number, then "nCr" and then the lower number, at least on most calculators I've seen.
No need to work with factorials then.

As for the Hypergeometric method, I thought the wikipedia page would explain that, but:

N = the total number of elements to choose from (30 here)
m = the number of elements in N that have whatever you are looking for (in this case the chocolates with rum = 5)
n = the amount of elements you select (in this case you want the first 16, so you select 16)
k = the number of elements with the attribute you're looking for that end up in n (here you want no rum chocolates in n, thus 0)

Excel is usually nice to use but since all we're usually allowed to use in exams are a pen and a calculator, I wouldn't rely on it if it's the same for you. Not yet anyway. ~;)

[Centurion1]

yeah moved past that bad boy hahaha!!!!

but if you have any insight on my second problem..... please fire away the answer eludes me. I tohught i had it with bernoulli but now i dont know...

[Montmorency]

Alright:

50C41.941.19
+
50C42.942.18
+
50C43.943.17
+
50C44.944.16
+
50C45.945.15
+
50C46.946.14
+
50C47.947.13
+
50C48.948.12
+
50C49.949.11
+
50C50.950.10

= egg

1 - egg = answer

I'm sure mathematics get much more difficult than this! Were you able to do the last problem? If so, post it: I bet I'll have no idea where to begin. :wacko:

[Centurion1]

i ended up getting that one.

i got stuck on this one though.

5. J.D. Power and Associates’ U.S. 2011 Wireless Call Quality Performance Study reports that the industry-average problem rate for smartphones is 13 per 100 calls. Leon is getting a new smartphone and calculates that he will make 50 calls this year.

b. If an exponential distribution applies for the number of calls between problems in this industry, what is the probability that Leon will not encounter any problems on the first half of the calls he will make this year? [10 pts.]