Question

Let X denote the number of bits received in error in a digital communication channel, and assume that X is a binomial with p = 0.0002. If 1000 bits are transmitted, determine the following. Round your answers to four decimal places (e.g. 98.7654).

(a) P(X = 1)

(b) P(X ≥ 1)

(c) P(X ≤ 2)

(d) mean and variance of X.

Answer 1

Answer:

a) 0.1638

b) 0.1813

c) 0.9989

d) Mean: $E(X) = 0.2$

Variance: $V(X) = 0.19996$

Step-by-step explanation:

For each bit received, there are only two possible outcomes. Either there is an error, or there is not. The probabilities of an error is independent for each bit. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The mean value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

In this problem we have that:

$n = 1000, p = 0.0002$

(a) P(X = 1)

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

$P(X = 1) = C{1000,1}*(0.0002)^{1}*(0.9998)^{999} = 0.1638$

(b) P(X ≥ 1)

We know that

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C{1000,0}*(0.0002)^{0}*(0.9998)^{1000} = 0.8187$

So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8187 = 0.1813$

(c) P(X ≤ 2)

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C{1000,0}*(0.0002)^{0}*(0.9998)^{1000} = 0.8187$

$P(X = 1) = C{1000,1}*(0.0002)^{1}*(0.9998)^{999} = 0.1638$

$P(X = 2) = C{1000,2}*(0.0002)^{2}*(0.9998)^{998} = 0.0164$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.8187 + 0.1638 + 0.0164 = 0.9989$

(d) mean and variance of X.

Mean

$E(X) = np = 1000*0.0002 = 0.2$

Variance

$V(X) = np(1-p) = 1000*0.0002*(1-0.0002) = 0.19996$