Answer: 0.171887
Step-by-step explanation:
Given that S0 and S1 are binary symbol of equal probabilty;
P(S0) = P(S1) = 0.5
This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.
P(Sn =X) = nCx p^x (1-p)^(n-1)
Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)
Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),...(1111111110) for Sn= 1 is the sequence for 1 error.
10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5
P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7
= 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10
=0.000977 + 0.00977 + 0.04395 + 0.11719
= 0.171887
