If a 10 symbol sequence is sent through the channel, what is the probability that up to 3 symbols are in error out of the 10transmissions. Show how to make a calculation to obtain a final value.

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