A coin is tossed $12$ times. How many different sequences of tosses could there have been, if the number of heads was at least $5?$
At least 5 heads means the same as no more than 4 tails
0 tails = C(12,0) = 1 possible sequence
1 tail = C(12,1) = 12 possible sequences
2 tails = C(12,,2) = 66 possible sequences
3 tails = C(12, 3) = 220 possible sequences
4 tails = C (12,4) = 495 possible sequences
At least 5 heads = sum of 12rh row of Pascal's Triangle - ( 1 + 12 + 66 + 220 + 495) =
2^12 - ( 1 + 12 + 66 + 220 + 495) =
3302 possible sequences