# Basic Probability Question

• Centurion1
In summary, 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 Thelmaf

## Homework Statement

1. 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?

## The Attempt at a Solution

Right so I know that the chances of getting a rum chocolate are at 17% to start off. And thne with each chocolate taken it decreases accordingly; i.e. 5/29, 5/28, etc. And that ely can take 16 chocolates before she goes over her diet

But I am not sure how to put it all together. I am not really looking for the blatant answer but more the process to find the answer. Obviously this is very basic probability and the actual math is not challenging.

You want to think about the probability of not choosing a rum chocolate. For her to consume 80% of her calorie limit, she has to choose 16 non-rum chocolates. What's the probability of doing so?

what is it 1.4% that you reach by multiplying it out?

Yes, that's it. Can you write it in terms of binomial coefficients?

well i actually did a hypergeometric.

but otherwise multiplying it out manually it is like

25/30 * 24/29 * 23/28 etc. until you do 16 trials.

im going to be asking another question in a different thread which i hope is okay?

well i actually did a hypergeometric.
Good! That's what I was trying to get at but I can never remember the name of the distribution.

Thank you for your help!