Probability of Getting Blue M&M from Mini Bag of 20

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Homework Help Overview

The discussion revolves around the probability of drawing a blue M&M from a mini bag containing 20 M&Ms, given that the first 19 drawn are blue. The stated distribution of colors includes 24% blue among other colors.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the probability of the 20th M&M being blue remains at 0.24, considering the independence of events and the implications of drawing multiple blue M&Ms.

Discussion Status

Some participants express uncertainty about the independence of the draws and the implications of the observed outcomes on the probability. Others suggest that the distribution provided should be assumed valid, while questioning the independence of the events based on the contents of the bag.

Contextual Notes

There is a mention of a potential misunderstanding regarding the independence of events when considering the fixed contents of the bag versus the probabilities given by the company. Additionally, a related question about calculating probabilities for combinations of different colored M&Ms is introduced, indicating a broader inquiry into probability concepts.

lopko
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Homework Statement


Lets say you have a mini bag of 20 M&Ms and find the first 19 are all blue. What is the probability that the 20th is also blue?

According to the company the distribution is 24% blue, 14% brown, 16% green, 20% orange, 13% red, 14% yellow

Thanks!



Homework Equations


I don't think this is relevant



The Attempt at a Solution


Would it just be the initial probability of getting a blue M&M (.24 according to the company), or does the fact that ALL of them in the bag are blue have an effect on the probability? I am assuming that the picking of each M&M are independent events, thus making the probability the same... but I am not sure if that's correct
 
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lopko said:

Homework Statement


Lets say you have a mini bag of 20 M&Ms and find the first 19 are all blue. What is the probability that the 20th is also blue?

According to the company the distribution is 24% blue, 14% brown, 16% green, 20% orange, 13% red, 14% yellow

Thanks!



Homework Equations


I don't think this is relevant



The Attempt at a Solution


Would it just be the initial probability of getting a blue M&M (.24 according to the company), or does the fact that ALL of them in the bag are blue have an effect on the probability? I am assuming that the picking of each M&M are independent events, thus making the probability the same... but I am not sure if that's correct

That would be correct.
 
Thanks Dick. What's the reasoning behind it?
 
lopko said:
Thanks Dick. What's the reasoning behind it?

They gave you the distribution of colors. And there's no reason not to assume the events are independent. The colors you've already seen don't affect the probability of the color of the next one. Of course, after pulling out 20 blues you might suspect the color distribution you've been given is wrong. But that really has nothing to do with this problem.
 
lopko said:

Homework Statement


Lets say you have a mini bag of 20 M&Ms and find the first 19 are all blue. What is the probability that the 20th is also blue?

According to the company the distribution is 24% blue, 14% brown, 16% green, 20% orange, 13% red, 14% yellow

Thanks!



Homework Equations


I don't think this is relevant



The Attempt at a Solution


Would it just be the initial probability of getting a blue M&M (.24 according to the company), or does the fact that ALL of them in the bag are blue have an effect on the probability? I am assuming that the picking of each M&M are independent events, thus making the probability the same... but I am not sure if that's correct

You need to be careful. The point is that the contents of the bag were fixed at the M&M factory, so the question is about the number of blues that were put into the bag at the plant. The probability 0.24 governs what was put into the bag, not necessarily what will be taken out be a customer. In particular, you need to worry about whether the successive colors drawn out of the bag by a customer are truly independent.
 
That's what I thought but I was involved in an argument here. Thanks.
I am confused about the question below. It is a rider to the one above.I believe it is a combination since order is not spoken of, but I am not sure.

If you are given 10 Brown M&Ms, 5 Yellow,16 Green, 7 Red, 11 orange and 6 blue, What would be FORMULA for the probability of obtaining ANY given combination of colors? Please define all variables used.
Sorry for being such an Oliver Twist.
 
Dick said:
They gave you the distribution of colors. And there's no reason not to assume the events are independent. The colors you've already seen don't affect the probability of the color of the next one. Of course, after pulling out 20 blues you might suspect the color distribution you've been given is wrong. But that really has nothing to do with this problem.

I think the question wants P{X = 20|X >= 19} in a binomial distribution X~Bin(20,0.24). That conditional probability is definitely not equal to 0.24. Further, P{X = k+1|X >=k} depends on k, so the results are not independent!
 
Last edited:
Ray Vickson said:
P{X >= k+1|X >=k} depends on k, so the results are not independent!
Ray, I don't think that's relevant. That's where you're looking at the total number of blues in the bag, not the first so many drawn at random. The logic in the OP looks right to me. Pulling the Mth M&M from the Bth bag of N M&Ms is equivalent to pulling the (B-1)*N+Mth from the production line.
 

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