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Let \({{\rm{p}}_{\rm{1}}}\) denote the probability that any particular code symbol is erroneously transmitted through a communication system. Assume that on different symbols, errors occur independently of one another. Suppose also that with probability \({{\rm{p}}_{\rm{2}}}\) an erroneous symbol is corrected upon receipt. Let \({\rm{X}}\) denote the number of correct symbols in a message block consisting of n symbols (after the correction process has ended). What is the probability distribution of \({\rm{X}}\)?

Short Answer

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Answer

The probability distribution of\({\rm{X}}\)is\({\rm{X}} \sim {\rm{Bin(n,1 - }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{1}}}{{\rm{p}}_{\rm{2}}}{\rm{)}}\).

Step by step solution

01

Concept Introduction

The amount of 'Success' in a series of n experiments, where each time a yes-no question is given, the Boolean-valued outcome is represented either with success/yes/true/one (probability\({\rm{p}}\)) or failure/no/false/zero (probability\({\rm{q = 1 - p}}\)) in a binomial probability distribution.

02

Probability Distribution

By description of the random variable \({\rm{X}}\), it follows Binomial Distribution where one of the parameters is \({\rm{n}}\) (can take values from \({\rm{0}}\)- not a single correct symbol in the message, up to \({\rm{n}}\) - all \({\rm{n}}\) symbols are correct).

The second parameter\({\rm{p}}\)needs to be calculated. Denote events –

\({\rm{Se = }}\){the symbol is erroneously transmitted};

\({\rm{Sc = }}\){the symbol is corrected upon receipt}.

The information given is –

\(\begin{array}{c}{\rm{P(Se) = }}{{\rm{p}}_{\rm{1}}}\\{\rm{P(Sc) = }}{{\rm{p}}_{\rm{2}}}\end{array}\)

The success in the Binomial Experiment would be event –

\({\rm{S = (Se)'}} \cup {\rm{(Se}} \cap {\rm{Sc),}}\)

Or in words this would mean that the symbol is correctly transmitted or the symbol is erroneously transmitted and corrected upon receipt. Event\({\rm{Se'}}\)and\({\rm{Se}} \cap {\rm{Sc}}\)are disjoint, therefore the following is true –

\(\begin{array}{c}{\rm{p = P(S) = P}}\left( {{\rm{Se'}} \cup {\rm{(Se}} \cap {\rm{Sc)}}} \right){\rm{ = P}}\left( {{\rm{Se'}}} \right){\rm{ + P(Se}} \cap {\rm{Sc)}}\\{\rm{ = 1 - P(Se) + P(Se)}} \cdot {\rm{P(Sc)}}\\{\rm{ = 1 - }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{1}}}{{\rm{p}}_{\rm{2}}}\end{array}\)

Finally, the probability distribution of \({\rm{X}}\) is –

\({\rm{X}} \sim {\rm{Bin(n,1 - }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{1}}}{{\rm{p}}_{\rm{2}}}{\rm{)}}\)

Therefore, the value is obtained as \({\rm{X}} \sim {\rm{Bin(n,1 - }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{1}}}{{\rm{p}}_{\rm{2}}}{\rm{)}}\).

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Most popular questions from this chapter

Ageologist has collected 10 specimens of basaltic rock and \({\rm{10 }}\) specimens of granite. The geologist instructs a laboratory assistant to randomly select \({\rm{15}}\) of the specimens for analysis.

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