# How to calculate the following probability?

I have a question that seems easy but kind of hard to calculate (at least for me):

There are three boxes of balls, one with red balls, one with blue balls, one with green balls. Draw three balls from one box, (so all three balls must be the same color) show them to 3 person P1 P2 P3 from distance and get their answers.

P1 P2 P3 have probabilities of correct identification P1=0.9 P2=0.8 P3=0.7

For the 1st ball, P1 says it is Blue, P2 says it is Blue, P3 says it is Red

For the 2nd ball, P1 says it is Blue, P2 says it is Green, P3 says it is Blue

For the 3rd ball, P1 says it is Red, P2 says it is Blue, P3 says it is Green

One thing in particlar that I don't understand is how to calculate the probability of all three balls being Blue? Is this equal to draw one ball and let P1 P2 P3 identify /measure three times?

• Please read the help center in relation to homework-style questions Sep 4, 2017 at 5:58
• The one thing in particular that I don't understand is that: "Draw three balls from one box". In other words, all three balls are the same color. So three balls, same color, measured by three people. Is this the same as one ball, measured by three people three times? Sep 4, 2017 at 8:25

To simplify notations, let B denote blue, R denote red and G denote Green.

The answers from P1, P2 and P3 is the observations (data), denoted as D.

The color state of the three balls (from which box were the three balls drawn) x is unknown and the question asks for the posterior probability mass function of x. E.g., $p(x = B|D)$ is the posterior probability of the three balls being Blue.

With the information provided, we should assign equal prior probability to all color states (uniform prior), i.e., $p(x = B) = p(x = G) = p(x = R)$.

Likelihood L(x|D) is the probability of P1, P2 and P3 giving the observed answers when the color states is x. E.g.,

$$L(B|D)=(0.9*0.8*0.3)*(0.9*0.2*0.7)*(0.1*0.8*0.3)$$

Similarly, you can calculate L(G|D) and L(R|D).

Then simply apply Bayes rule, e.g.,

$$p(B|D) = \frac{L(B|D)p(B)}{\sum_{x}L(x|D)p(x)}=\frac{L(B|D)}{L(B|D)+L(G|D)+L(R|D)}.$$

If the question changes from "draw three balls from one box" to "draw one ball from each box" and asks for, say, the probability of 1st ball being Blue, the 2nd ball being Red and the 3rd ball being Green, one can use similar procedure to solve the question with a slightly more complicated set of states:

The color states of the three balls (note that the ordering matters here) xyz is unknown and the question asks for the posterior probability mass function of xyz. E.g., $p(xyz = BRG|D)$ is the posterior probability of the 1st ball being Blue, the 2nd ball being Red and the 3rd ball being Green.

• Thank you very much for your answer. However, it states "Draw three balls from one box", so all three balls are from the same box and they must be the same color. When I try to calculate L(BBB |D) from your answer, it seems not right. Would you please help me out? Sep 4, 2017 at 8:19
• answer modified as requested; L(BBB |D) (or L(B|D) in the new version) is not the answer, you need to apply Bayes rule Sep 4, 2017 at 12:44
• Thank you again, I got where I didn't calculate right, I used L(x|D) instead of L(x|D)p(x). Sep 4, 2017 at 20:49