Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Question: A space probe near Neptune communicates with Earth using bit strings. Suppose that in its transmissions it sends a 1 one-third of the time and a 0 two-thirds of the time. When a 0 is sent, the probability that it is received correctly is 0.9, and the probability that it is received incorrectly (as a 1) is 0.1. When a 1 is sent, the probability that it is received correctly is 0.8, and the probability that it is received incorrectly (as a 0) is 0.2
a) Find the probability that a 0 is received.
b) Use Hayes theorem to find the probability that a 0 was transmitted, given that a 0 was received.

Answer:

Probability that 0 received is \( = \frac{2}{3}\)

Probability that 0 transmitted and 0 is received \( = 0.9\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Probability that 1 sent is received correctly = 0.8

Probability that 1 sent is received incorrectly = 0.2

Step 2: Formula used

\(p\left( {\frac{E}{{{E_1}}}} \right) = \frac{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right)}}{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right) + p\left( {\frac{{{E_2}}}{F}} \right)p\left( F \right)}}\)

Step 3: Calculating (a)

In first part we need to find the probability that a 0 is received at earth. This can be due to 2 reasons

0 is sent and 0 is received
1 is sent and 0 is received

Thus,

\(p\left( {0\,is sent and 0 is received} \right) = \frac{2}{3} \times 0.9\)

\(p\left( {1\,\,is sent and 0 is received} \right) = \frac{1}{3} \times 0.2\)

\(\begin{array}{l}p\left( 0 \right) = \frac{2}{3} \times 0.8 + \frac{2}{3} \times 0.2\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{3}\end{array}\)

Thus, the final probability is 2/3

Step 4: Calculating (b)

The probability that 0 is received given that 0 is sent is simply the probability that 0 is received correctly as we have already established that 0 is sent.

\(p\left( {0\,is\,recieved\,given\,that\,0\,is\,sent} \right) = 0.9\)

\(p\left( {0\,\,is\,sent} \right) = \frac{2}{3}\)

\(p\left( {0\,\,is\,received} \right) = \frac{2}{3}\)(from part 1)

Probability that 0 is sent is 2/3 due to the equipment. The remaining probability is taken from the answer of part 1

\(p\left( {0\,\,is\,sent\,given that 0 is recieved} \right) = 0.9 \times \frac{{2/3}}{{2/3}}\)

\( = 0.9\)

The probability that 0 was transmitted given that a 0 was received is 0.9

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