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I don\'t understand this please explain how you get your answer, and please make

ID: 3048233 • Letter: I

Question

I don't understand this please explain how you get your answer, and please make a tree diagram because I am also unsure how to label the diagram.

1. When an engineer wants to send a binary signal (a 0 or 1 signal) through a communication channel, assume that there is a 10% chance that the wrong signal is received, that is, if the transmitted signal is a 1, then there is a 10% chance that the signal received is a 0; while if the transmitted signal is a 0, then there is a 10% chance that the signal received is a 1. In order to improve the signal transmission reliability, the signal to be transmitted will be sent five independent times to the receiver. The receiver will then interpret the signal to be a 1 if he/she receives at least three 1’s among the five transmissions, while he/she will interpret the signal to be a 0 if he/she receives at least three 0’s among the five transmissions.

(a) With this system, what is the probability of getting the correct signal? [This probability is called the reliability of the communication or transmission system.] Is the probability high enough?

(b) Suppose you wanted to achieve at least a 0.999999 reliability for your communication system. Denote by n the odd number of times that you will be transmitting your signal (in the preceding we took n = 5), and assume that you will interpret the signal to be a 1 if your receive at least (n + 1)/2 1’s among the n transmitted signals, and 0 otherwise. What is the appropriate value of n?

Explanation / Answer

Tree diagram is not needed in this case. It can be solved using binomial.

a) P( 1 to be received if 1 is sent) =p= 0.9

P(1) =Probabilty that 1 is received if 1 is sent 3,4or 5 times.= P(3)+P(4)+P(5)= 5C3 *0.9^3*0.1^2+5C4*0.9^4*0.1+5*0.9^5=10*0.9^3*0.1^2+5*0.9^4*0.1+1*0.9^5=0.99144

b) n = 25 is found out using a binomal calculator.

We calculate binomal probabilites for 11,13,15....and so on till we find that P=0.999999 and above