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 0 is sent as(000). What is the probability that the receiver assigns the correct value of 0?

Answer 1

Answer:

• The probability that the receiver assigns the correct value of 0 is 0.017.

Explanation:

As per the description, there is a typo in the final statement. The correct final statement has to be "A is sent as 00000 " beacuse each digit is sent five times.

The receiver will assign the correct value of 0 if 0 appears three or more times.

Then, you need to find the probability that the string contains 3 or more 0.

Since the variable can take two values (0 or 1) and the probability of each bit are independent this is a binomial experiment.

Thus, the calculation is to find P(X≥3), which is equal to P(X=3) + P(X=4) + P(X=5).

The equation for the binomial probability is:

          $P(X=x)=C_{n,x}p^x(1-p)^n=\frac{n!}{x!(n-x)!}p^x(1-p)^{(n-x)}$

For this experiment, p is the probability that the bit is not reversed, thus p = 0.7, and (1-p) = 0.3.

Then, you need to find:

      $C_{5,3}(0.7)^3(0.3)^2+C_{5,4}(0.7)^4(0.3)^1+C_{5,5}(0.7)^5(0.3)^0$

Computing, that is 0.01699 ≈ 0.017