5.3 MM's

From various sources I have found the fraction of chocolate M&Ms candies are red. The sources found are the following:

Source A: 28% of M&Ms are red, 20% of M&Ms are orange.
Source B: 20% of M&Ms are red, 10% of M&Ms are orange
Source C: 13% of M&Ms are red, 21% of M&Ms are orange.

From actually counting of a bag of M&Ms I found the following data:

3 red M&Ms in 17 total ( $R = 3, N = 17$ )

The question is, which source can we trust the most? Here we follow Bayes' recipe,

Try it in XYZ Crunch

Each source's likelihood is one binomial lookup

Every likelihood on this page is the binomial probability of exactly 3 red in 17, at that source's claimed fraction. The panel gives P(X <= k), so subtract consecutive lookups and then divide by 3 for the equal prior: Source A 0.256011 - 0.105823 = 0.150189, over 3 = 0.05006. Source B 0.548876 - 0.309622 = 0.239254, over 3 = 0.07975. Source C 0.828995 - 0.616373 = 0.212622, over 3 = 0.07087. Those are the three numbers the book sums to K = 0.20068.

Specify the prior probabilities for the models being considered
$\begin{matrix} P (A) = P (B) = P (C) = 1 / 3 \end{matrix}$
Write the top of Bayes' Rule (i.e. likelihood $\times$ prior) for all models being considered
$\begin{matrix} P (A | R = 3, N = 17) \sim (\binom{17}{3}) {0.28}^{3} (1 - 0.28)^{17 - 3} \times \frac{1}{3} \\ P (B | R = 3, N = 17) \sim (\binom{17}{3}) {0.20}^{3} (1 - 0.20)^{17 - 3} \times \frac{1}{3} \\ P (C | R = 3, N = 17) \sim (\binom{17}{3}) {0.13}^{3} (1 - 0.13)^{17 - 3} \times \frac{1}{3} \end{matrix}$
Add these values for all models, to get $K$
$\begin{matrix} P (A | R = 3, N = 17) \sim 0.05006 \\ + \\ P (B | R = 3, N = 17) \sim 0.07975 \\ + \\ P (C | R = 3, N = 17) \sim 0.07087 \\ K = 0.20068 \end{matrix}$
Divide each of the values by this sum, $K$ , to get the final probabilities
$\begin{matrix} P (A | R = 3, N = 17) = 0.05006 / 0.20068 = 0.250 \\ P (B | R = 3, N = 17) = 0.07975 / 0.20068 = 0.397 \\ P (C | R = 3, N = 17) = 0.07087 / 0.20068 = 0.353 \end{matrix}$

So we are most confident in Source B, although none of them really changed by a lot - there is no clear winner.

Updating with other data

5 orange M&Ms in 16 total ( $G = 5, N = 16$ )

Again, we follow the same recipe, starting with out posterior probabilities from above as our starting prior probabilities - they are prior to the new data.

Specify the prior probabilities for the models being considered
$\begin{matrix} P (A | o l d d a t a) = 0.250 \\ P (B | o l d d a t a) = 0.07975 / 0.20068 = 0.397 \\ P (C | o l d d a t a) = 0.07087 / 0.20068 = 0.353 \end{matrix}$
Write the top of Bayes' Rule (i.e. likelihood $\times$ prior) for all models being considered
$\begin{matrix} P (A | G = 5, N = 16 and old data) \sim (\binom{16}{5}) {0.20}^{5} (1 - 0.20)^{16 - 5} \times 0.250 \\ P (B | G = 5, N = 16 and old data) \sim (\binom{16}{5}) {0.10}^{5} (1 - 0.10)^{16 - 5} \times 0.397 \\ P (C | G = 5, N = 16 and old data) \sim (\binom{16}{5}) {0.21}^{5} (1 - 0.21)^{16 - 5} \times 0.353 \end{matrix}$
Add these values for all models, to get $K$
$\begin{matrix} P (A | d a t a) \sim 0.0300 \\ + \\ P (B | d a t a) \sim 0.00544 \\ + \\ P (C | d a t a) \sim 0.0471 \\ K = 0.08254 \end{matrix}$
Divide each of the values by this sum, $K$ , to get the final probabilities
$\begin{matrix} P (A | d a t a) = 0.0300 / 0.08254 = 0.363 \\ P (B | d a t a) \sim 0.00544 / 0.08254 = 0.0659 \\ P (C | d a t a) \sim 0.0471 / 0.08254 = 0.5706 \end{matrix}$

Given this new data, we update our state of knowledge, and we're much more confident that Source C is the best one. It is clear that Source B is unlikely, with a probability of only about 6.5%. We could extend this example with more data, and more models if we'd like.

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.