More subsets and counting #combinatorics #discretemath #math #tutor #amc #what #why #mybrainhurts

I think this is correct: the number of total subsets = 2^7. We assume the subset has 1 and 2 in it, so we are only left with choosing if the subset contains any of 3 through 7, which is 5 choices. Aka, we have 2^5 possible choices for subsets with 1 and 2 in it. Thus, the probability is 2^5 / 2^7 = 1/4.

1/4.

For all the permutations of the elements of 3-7, there exists a subset with the null set, with 1, with 2, and with 1,2.

So for every subset that fits the criteria, there are 3 more. Ergo, 1/4 of all subsets fit the criteria.