[Actually you can’t be dickish/clever that way: The problem isn’t underspecified as the goal is to do the best you can with the information you’ve got. You’ve got no information/evidence regarding the distribution between classes so your best bet is to treat it as random. From there you can use Bayes theorem, blahblah, etc.etc....]
In words, this says that to generate the i-th class, you flip a coin to tell whether it’s in program A or program B, conditioned on the program, the proportion of boys is drawn from a program-specific beta distribution, and then the number of boys is drawn from the corresponding binomial distribution. Under the constraints that
%20=%200.65) and
%20=%200.45), the average proportion of boys matches up with the problem.
However, by taking a_0 or b_0 small (where a_1 and b_1 are adjusted accordingly to maintain the constraint), you can play with the variance so that the observed 55% boys class is more likely under either of the programs. If you had available repeated trials, you might be able to learn a_0 and b_0. In a single trial, you can’t be sure that your strategy will do worse than chance.
[Actually you can’t be dickish/clever that way: The problem isn’t underspecified as the goal is to do the best you can with the information you’ve got. You’ve got no information/evidence regarding the distribution between classes so your best bet is to treat it as random. From there you can use Bayes theorem, blah blah, etc. etc....]
Or just change the 45% and 65% to 11% and 99%. That makes the correct answer pretty obvious without changing anything important.
Oops, you’re right. The variant of the problem I mentioned above got rid of the assumption of binomially distributed boys (equivalently, girls).
The following setup should work, though:
In words, this says that to generate the i-th class, you flip a coin to tell whether it’s in program A or program B, conditioned on the program, the proportion of boys is drawn from a program-specific beta distribution, and then the number of boys is drawn from the corresponding binomial distribution. Under the constraints that
However, by taking a_0 or b_0 small (where a_1 and b_1 are adjusted accordingly to maintain the constraint), you can play with the variance so that the observed 55% boys class is more likely under either of the programs. If you had available repeated trials, you might be able to learn a_0 and b_0. In a single trial, you can’t be sure that your strategy will do worse than chance.