Consider a binary code with 4 bits (0 or 1) in each code word. In each

siroticuvj

siroticuvj

Answered question

2021-11-21

Think about a binary code that uses 4 bits (0 or 1) per code word. A bit in every code word is zero with a probability of 0.6, unrelated to any other bit.
(a) What is the probability of the code word 0101? 
(b) What is the probability that a code word contains exactly two ones?

Answer & Explanation

Coldst

Coldst

Beginner2021-11-22Added 18 answers

Step 1 
Given data: 
The probability of 0 bit is P(0)=0.6
The probability of one bit is expressed as,
P(1)=1P(0) 
Replace the specified values in the aforementioned expression.
P(1)=10.6 
=0.4 
a) The expression for the probability of code-word 0101 is, 
P(a)=P(0)×P(1)×P(0)×P(1) 
Substitute the given and above-calculated values in the expression. 
P(a)=(0.6)(0.4)(0.6)(0.4) 
=0.0576 
Thus, the probability of code-word 0101 is 0.0576. 
Step 2 
b) The expression for the probability that the code-word contains exactly 2 ones is, 
P(b)=4C2(P(1))2(P(0))2 
Substitute the given values in the above expression. 
P(b)=4C2(0.4)2(0.6)2 
=4!(2!)(2!)(0.4)2(0.6)2 
=6×(0.4)2(0.6)2 
=0.3456 
Thus, the probability that the code-word contains exactly 2 ones is 0.3456.

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