This is a binomial distribution and the answer depends on the total number of Type A and B questions as well as what confidence level you want to use for the cutoff. There is probably a formula for determining what you want, but I don't know what it is. You can compute it yourself using a permutation test.
Generate a null hypothesis distribution of random guesses and then use this to find the cutoff for whatever p-value (probability that randomly guessing will give a score greater than a certain value) you are comfortable with. The easiest way to get that value is to sort all of the null performance scores and get one that is 95% higher than all of the others (say Score #950 if you did 1000 random guesses and sorted the scores).
Here is some Matlab code to estimate this:
n_trials = 50;
p_cutoff = 0.05;
n_permutations = 100000;
% Create a vector of n_trials 'correct' responses and replicate it across n_permutations
expected_values = repmat(randi([0 1],[1 n_trials]),[n_permutations 1]);
% Create an array of n_permutations x n_trials random guesses
random_responses = randi([0 1],[n_permutations n_trials]);
% Compute the percent correct for each permutation
percent_correct = sum(expected_values==random_responses,2)/n_trials;
% Compute and print the cutoff percentage for desired p value
fprintf('\np=%1.3f Cutoff for %d trials: %3.1f%%\n\n', p_cutoff, n_trials, quantile(percent_correct,1-p_cutoff)*100);
Here are some cutoffs for p=0.05 assuming the same number of Type A and Type B questions:
More total questions (trials) and the cutoff gets closer to 50%.
Note this is all assuming that there is no kind of bias, say a participant choosing randomly will be more likely to choose A (like the first answer or something like that).
Here is an online calculator for computing some probabilities with the binomial distribution.