Problem: a fair coin is tossed 11 times . what is the probability of getting more heads than tails in 11 tosses ?
Rationale: "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ."
annotated_formula: divide(add(add(add(choose(11, const_2), choose(11, const_3)), choose(11, const_4)), choose(11, 11)), power(const_2, 11))