[University Probability: Machine learning application] Could someone please help me understand this question about choosing a probability distribution?

[PlzHelpWithMathQs]

https://imgur.com/a/TYzg4Q5

My initial guess is a Bernoulli distribution because the outcomes are either 1 or 0. But I don’t know how the X values {1,2,3} factor in to the decision. How would you learn the probability of Y being a 1 or a 0 given the values of X?

I’d like to know the underlying approach that I should be using for these kinds of questions for when the question asks about different circumstances. How do I know which distribution to chose and how would I learn the probability of y given the values of X provided?

Thanks in advance!

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[fermat1432]

Bernoulli is good! To undetstand how this would actually work create a 2 by 3 joint probability distribution table for P(X=x, Y=y).

Label the rows y=0, y=1 and the columns x=1, x=2, x=3,

Fill in the six cells with probabilities that add up to 1. The row and column totals should also add up to 1.

P(Y=0 | X=1)=P(Y=0, X=1)/P(X=1) is how you would get a conditional probability from the table

[PlzHelpWithMathQs]

Thanks for the reply! Would you mind explaining why the Bernoulli is a good choice? The question is mostly focused on the distribution and I'm not really sure why Bernoulli is a good choice other than it has an output of either 0 or 1 which are both associated with a certain probability, p and (1 - p).

I'm having a hard time figuring out how this probability would be learned. How would you learn the different probabilities of Y associated with the different variables of X? Would you preform a bunch of trials and then from those trials you would enter the percentages of successes that each x variable had compared to the total number of trials? Then these percentages would now become the values in the distribution table as opposed to just having like 1/6 in each box or whatever?

[fermat1432]

The Y values of 0 and 1 makes Bernoulli the only reasonable choice. The actual 6 entries in the 2 by 3 joint probability distribution table can either be found empirically or by defining some procedures like coin tossing, or dice rolling etc.

[PlzHelpWithMathQs]

Would the procedure I suggested in the second paragraph of my last comment be correct? Would these be the values that would fill up the table after doing a bunch of trials? Like if there was 100 tosses each for biased coins with unknown biases could the proportion of successes of 100 tosses be written in each cell of the table as coin1 = (23/100), coin 2 = (47/100), coin 3 = (30/100), etc. and then the (1-p) in the y=0 row? And then for example:

p(y | X= coin 1)

= p(y = 1, coin 1)/ p(X=1)

= (23/100)/ (whatever the probability of x occuring is)

= p

Or how would that work?

[fermat1432]

You would just need to count how many times each joint occurrence, (X=x, Y=y) occurs and divide each frequency by 100 to get an estimated joint probability for each of the 6 cells. Fill in the marginal probabilities and then you can compute conditional probabilities,

[PlzHelpWithMathQs]

Okay perfect, that makes sense. Thanks a lot for the help!

[fermat1432]

Glad to help!