Question

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?

Answer 1

Answer:

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%)

Answer 2

The probability of a code phrase containing exactly one zero is calculated below.

Suppose X refers to the number of 0's in binary digits of 5 bits (0 or 1) per codeword.

$\to X \sim Binomial(5, 0.8)$

X has a pmf:

$\to P(X=x)=\binom{5}{x}\times 0.8^x \times 0.2^{5-x}, \ \ x=0,1,2,3,4,5$

 The probability that a code word contains exactly one zero:

$\to P(X=1)=\binom{5}{x} \times 0.8^1 \times 0.2^{5-1}=0.0064$

Therefore, the final answer is "0.0064".

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