We flip a fair coin 10 times. What is the probability that we get heads in at least 8 of the 10 flips?

P (Heads  ≥ 8 flips)  =  P( Tails ≤ 2 flips )  =

C(10,0)(1/2)^10   + C(10,1) (1/2)^10  + C(10,2)(1/2)^10  =  7 / 128

The 8th term of tetranacci sequence are the odds out 2^10 chances. Since the 8th term of this sequence is 56, therefore the odds of at least 8 Heads are: 56/(2^10)=0.0546875 ~ 5.5%.