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

siroticuvj 2021-11-21 Answered
Consider a binary code with 4 bits (0 or 1) in each code word. In each code word, a bit is a zero with probability 0.6, independent of 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?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Coldst
Answered 2021-11-22 Author has 751 answers
Step 1
Given data:
The probability of 0 bit is \(\displaystyle{P}{\left({0}\right)}={0.6}\).
The expression for the probability of 1 bit is,
\(\displaystyle{P}{\left({1}\right)}={1}-{P}{\left({0}\right)}\)
Substitute the given values in the above expression.
\(\displaystyle{P}{\left({1}\right)}={1}-{0.6}\)
\(\displaystyle={0.4}\)
a) The expression for the probability of code-word 0101 is,
\(\displaystyle{P}{\left({a}\right)}={P}{\left({0}\right)}\times{P}{\left({1}\right)}\times{P}{\left({0}\right)}\times{P}{\left({1}\right)}\)
Substitute the given and above-calculated values in the expression.
\(\displaystyle{P}{\left({a}\right)}={\left({0.6}\right)}{\left({0.4}\right)}{\left({0.6}\right)}{\left({0.4}\right)}\)
\(\displaystyle={0.0576}\)
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,
\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({P}{\left({1}\right)}\right)}^{{{2}}}{\left({P}{\left({0}\right)}\right)}^{{{2}}}\)
Substitute the given values in the above expression.
\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)
\(\displaystyle={\frac{{{4}!}}{{{\left({2}!\right)}{\left({2}!\right)}}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)
\(\displaystyle={6}\times{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)
\(\displaystyle={0.3456}\)
Thus, the probability that the code-word contains exactly 2 ones is 0.3456.
Have a similar question?
Ask An Expert
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-04-06
Assign a binary code in some orderly manner to the 52 playingcards. Use the minimum number of bits.
asked 2021-09-22
a simple binary communication channel carries messages by using only two signals, say 0 and 1. We assume that, for a given binary channel, 40% of the time a | is transmitted. the probability that a transmitted 0 is correctly received is 0.90 and the probability that a transmitted | is correctly received is 0.95. given a 0 is received, the probability that 0 was transmitted.
asked 2021-09-14
a simple binary communication channel carries messages by using only two signals, say 0 and 1. We assume that, for a given binary channel, 40% of the time a | is transmitted. the probability that a transmitted 0 is correctly received is 0.90 and the probability that a transmitted | is correctly received is 0.95. given a | is received, the probability that 1 was transmitted.
asked 2021-09-13
Ali an Sara each choose a number independently and uniformly at random from interval [0,2].
Consider the following events:
A:The absolute difference between the two numbers is greater than 1/4
B: Ali`s number is greater than 1/4
Find the probability \(\displaystyle{P}{\left[{A}\cap{B}\right]}\)
asked 2021-09-20
a simple binary communication channel carries messages by using only two signals, say 0 and 1. We assume that, for a given binary channel, 40% of the time a | is transmitted. the probability that a transmitted 0 is correctly received is 0.90 and the probability that a transmitted | is correctly received is 0.95. the probability of a 1 being received
asked 2021-11-18
Assume the number of commuters using the QR code payment service at each MTR station follows a Poisson probability distribution. Based on a recent statistics, on average, 3 commuters use the QR code payment service in an hour.
(i) What is the probability there are exactly 10 commuters who use the QR code payment service in an hour?
(ii) What is the probability that there are less than 4 commuters who use the QR code payment service in 3 hours?
(iii) If five MTR stations are randomly selected, what is the probability that at least two of the five MTR stations will have less than 4 commuters who use the QR code payment service in 3 hours?

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...