A message consists of a string of bits (0s and1s). 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 receiverassigns 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?

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(X2) |

0 | 0.16807 |

1 | 0.36015 |

2 | 0.3087 |

Total | 0.83692 |

Therefore,

Step3 of 3:

b).

We have p = 0.3 and we need to find the smallest odd value of n.

X B(n, 0.3)

, x = 1, 3, 5, 7, 9. . . . . (1)

1). Put n = 3 in above equation(1) we get

is obtained from Excel by using the function “=binomdist(X,n,p,false)”

...X |