Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01001. In each code word, a bit is zero with probability 0.8, independent of any other bit. What is the probability that a code word contains exactly one zero?

the probability that a code word contains exactly one zero is 0.0064 (0.64%)

Step-by-step explanation:

Since each bit is independent from the others , then the random variable X= number of 0 s in the code word follows a binomial distribution, where

p(X)= n!/((n-x)!*x!*p^x*(1-p)^(n-x)

where

n= number of independent bits=5

x= number of 0 s

p= probability that a bit is 0 = 0.8

then for x=1

p(1) = n*p*(1-p)^(n-1) = 5*0.8*0.2^4 = 0.0064 (0.64%)

therefore the probability that a code word contains exactly one zero is  0.0064 (0.64%)

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see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph