Six fair coins are tossed at the same time. What is the probability of getting exactly four heads?

Total outcomes = $2^6=64$. The four heads can sit on any 4 of the 6 coins: $C(6,4)=15$. Probability = 15/64. Trap $11/32=(15+6+1)/64$ answers 'at least four heads' by piling on the five-head and six-head cases. Trap $5/16=C(6,3)/64$ answers 'exactly three heads', the half-and-half split. Trap 1/64 is 'all six heads'.