Problem: a fair coin is tossed 4 times . what is the probability of getting at least 2 tails ?
Rationale: "let ' s find the probability of the opposite event and subtract this value from 1 . the opposite event would be getting zero tails ( so all heads ) or 1 tail . p ( hhhh ) = ( 12 ) 4 = 116 p ( hhhh ) = ( 12 ) 4 = 116 . p ( thhh ) = 4 ! 3 ! ∗ ( 12 ) 4 = 416 p ( thhh ) = 4 ! 3 ! ∗ ( 12 ) 4 = 416 , we are multiplying by 4 ! 3 ! 4 ! 3 ! since thhh scenario can occur in number of ways : thhh , hthh , hhth , or hhht ( notice that 4 ! 3 ! 4 ! 3 ! basically gives number of arrangements of 4 letters thhh out of which 3 h ' s are identcal ) . p ( t ≥ 2 ) = 1 − ( 116 + 416 ) = 1116 p ( t ≥ 2 ) = 1 − ( 116 + 416 ) = 1116 . answer : d ."
annotated_formula: divide(add(add(add(choose(4, const_2), choose(4, const_3)), choose(4, const_4)), choose(4, 4)), power(const_2, 4))