Evaluation of sample space (Bayes' Theorem)

Question

This question is solved one - but I am trying to find flaw in my understanding of the situation. The question is as follows:

The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% Red, 10% Green, 10% Orange, 10% Tan). Afterward it was (24% Blue , 20% Green, 16% Orange, 14% Yellow, 13% Red, 13% Brown).

A friend of mine has two bags of M&Ms, and he tells me that one is from 1994 and one from 1996. He won't tell me which is which, but he gives me one M&M from each bag. One is yellow and one is green. What is the probability that the yellow M&M came from the 1994 bag?

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I am trying to find the sample space first, and it seems there is flaw in my understanding of the problem. Need help in understanding that flaw and if possible please let me know how you would solve it (with or without Bayes' Theorem). I am a beginner (not a student though) and want to understand different aspects from which this problem can be viewed. 2-3 approaches should suffice. I know that this will call extra pain - I want to thank you in advance.

Let B1 (1994) and B2 (1996) are two bags. The sample space should comprise of following events -

1. B1 x B2 = 36 events - That is an ordered pair containing first M&M from B1 and the other from B2.
2. B2 x Tan (from B1) = 6 events
3. B1 x Blue (from B2) = 6 events

So in total we have 48 possible events. (Please confirm)

However since the weight of each event varies because of difference in frequency of each colour the probability of each of these 48 events is not equally likely. Sum of the probability of all these events should be unity though.

P(Y/B1) . P(G/B2) = .04

What should I do from here on ?

Answer 1

Few points to consider before actually working out the solution.

1. We are not told about which bag is which. This is an important point.

2. For every pair there could be two possibilities. 1st color may come from 1994 bag or from 19996 bag. (x,y) or (y,x)

3. We don't know which color came from where. We have to find the numerical value for the possibility.

Evaluating Sample Space

There are six colors in each set and thus we have 36 different possibilities. Points 2 and 3 in the question do not matter because the order does not matter.

B1 x B2 = (b1 from B1 and b2 from B2) All colors are common in both bags except two, viz Tan and Blue.

Since Tan is only present in B1, only (Tan, b2) is possible similarly only (b1,Blue) is possible and not the reverse.

Apply Conditional probability (Bayes' Theorem)

The friend gave me a yellow and green M&M and there are two ways of getting them. (Y,G) and (G,Y). I don't know which is from where. Since I am only looking for yellow from 1994 bag, there are two different events as to how bags can be chosen.

E1 = B1, the first bag from which first M&M was drawn was 1994 bag and the other one was from 1996.

E2 = B1, the first bag from which first M&M was drawn was 1996 bag and the other one was from 1994.

In addition let I be the event of us getting a yellow M&M and a green M&M irrespective of choice of bags.

Now, we apply Bayes' Theorem as below.

P(E1 | I ) = P(I ∣ E1).P(E1) / P(I) ......Eq (1)

Evaluating P(I ∣ E1) we get .04 Evaluating P(E1) we get .5

The denominator of Eq. (1) is a harder than above ones.

P(I) = P(I ∩ (E1 ∪ E2))= P[(I ∩ E1) ∪ (I ∩ E2)]

= P[(I ∩ E1) ∪ (I ∩ E2)] =

= P (E1). P(I ∣ E1) + P (E2). P(I ∣ E2)

Now filling in the values here :

= (1/2). (.04) + (1/2).(0.014)

= 0.027

To compute the final answer it is just matter of substituting final values in Eq (1)

P(E1 | I ) = (0.04) (1/2) / (0.027)

= 0.741 (approx)

Answer 2

You don't have 48 combination events, just 36, since there are only 6 colour options in each case. Of course blue or tan will not have a reverse case to compare with, giving certainty. Alternatively, to preserve symmetry you could have 49 options across the total of 7 colours, with zeroes as appropriate for blue in older and tan in newer bags.

You do not need to calculate out the whole grid though. You can be mathematically certainly that the distributive law holds, giving a net sum of $1$. So just calculate the probabilities for $(yellow, green)$ and $(green, yellow)$ options and work from there. Calculating those probabilities is where any non-random selection process, noted in comments, will take effect.