Solution Found!
A message consists of a string of bits (0s and1s). Due to
Chapter 4, Problem 21E(choose chapter or problem)
A message consists of a string of bits (0s and 1s). Due to noise in the communications channel, each bit has probability 0.3 of being reversed (i.e., a 1 will be changed to a 0 or a 0 to a 1). To improve the accuracy of the communication, each bit is sent five times, so, for example, 0 is sent as 00000. The receiver assigns the value 0 if three or more of the bits are decoded as 0, and 1 if three or more of the bits are decoded as 1. Assume that errors occur independently.
a. A 0 is sent (as 00000). What is the probability that the receiver assigns the correct value of 0?
b. Assume that each bit is sent \(n\) times, where \(n\) is an odd number, and that the receiver assigns the value decoded in the majority of the bits. What is the minimum value of \(n\) necessary so that the probability that the correct value is assigned is at least 0.90?
Equation Transcription:
Text Transcription:
n
Questions & Answers
QUESTION:
A message consists of a string of bits (0s and 1s). Due to noise in the communications channel, each bit has probability 0.3 of being reversed (i.e., a 1 will be changed to a 0 or a 0 to a 1). To improve the accuracy of the communication, each bit is sent five times, so, for example, 0 is sent as 00000. The receiver assigns the value 0 if three or more of the bits are decoded as 0, and 1 if three or more of the bits are decoded as 1. Assume that errors occur independently.
a. A 0 is sent (as 00000). What is the probability that the receiver assigns the correct value of 0?
b. Assume that each bit is sent \(n\) times, where \(n\) is an odd number, and that the receiver assigns the value decoded in the majority of the bits. What is the minimum value of \(n\) necessary so that the probability that the correct value is assigned is at least 0.90?
Equation Transcription:
Text Transcription:
n
ANSWER:Solution 21E
Step1 of 3:
Let us consider a random variable X it presents the number of bits that are reserved.
Then X follows binomial distribution with parameters “n and p” that is X B(n, p),
The probability mass function of binomial distribution is given by
, x = 0,1,2,...,n.
Where,
n = sample size
= 5
x = random variable
p = probability of success
= 0.3
q = 1 - p (probability of failure)
= 1 - 0.30
= 0.70
Here our goal is:
a).We need to find the probability that the receiver assigns the correct value of 0.
b).We need to find the minimum value of n necessary so that the probability that the correct value is assigned is at least 0.90.
Step2 of 3:
a).
We have n = 5, p = 0.3 and correct value is assigned if
Then
Here can be obtained from Excel by using the function “=binomdist(X,n,p,false)”
X |
P(X |
0 |
0.16807 |
1 |
0.36015 |
2 |
0.3087 |
Total |
0.83692 |
Therefore,
Step3 of 3: