SOLUTION: The communication channel transmits a three-digit number M = 873, written in decimal digits. Each digit is passed correctly with probability 0.62 and is converted to any other digi

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Question 1192270: The communication channel transmits a three-digit number M = 873, written in decimal digits. Each digit is passed correctly with probability 0.62 and is converted to any other digit with probability (1 - 0.62) / 9.
a) what is the probability that transmitting the number M will decrease it?
b) what is the maximum increase in the probability of a decrease in the number when adding a digit from the right?

Answer by ikleyn(52834)   (Show Source): You can put this solution on YOUR website!
.
The communication channel transmits a three-digit number M = 873, written in decimal digits.
Each digit is passed correctly with probability 0.62 and is converted to any other digit
with probability (1 - 0.62) / 9.
a) what is the probability that transmitting the number M will decrease it?
b) what is the maximum increase in the probability of a decrease in the number when adding a digit from the right?
~~~~~~~~~~~~~~


                    I will solve part  (a),  only.


The probability that transmitting the number M= 873 will increase it is this sum


    P =   P(first digit 8 is no change)*P(second digit 7 is no change)*P(third digit 3 increases) +

        + P(first digit 8 is no change)*P(second digit 7 increases) +

        + P(first digit 8 is increases)  = 

        = 0.62*0.62*(6*((1-0.62)/9)) + 0.62*(2*((1-0.62)/9)) + 1*((1-0.62)/9) = 0.191959  (rounded).    ANSWER


It requires some comments.


P(third digit 3 increases) = 6*((1-0.62)/9),  since there are 6 possibilities to increase 3: they are 4,5,6,7,8,9.

P(second digit 7 increases) = 2*((1-0.62)/9), since there are 2 possibilities to increase 7: they are 8,9.

P(first digit 8 is increases) = 1*((1-0.62)/9), since there is only 1 possibility to increase 8: it is 9.

Solved.



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