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/wikiped...o_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.

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.