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.

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

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x