Question

When a message is sent electronically it is usually sent as a stream of bits, each of which can be either a 0 or a 1. If the digital channel is noisy then each
bit has some probability of being flipped (ie changed from a 0 to a 1 or vice versa) resulting in a corrupted message.
Assume that a message is being sent through a noisy channel where the probability that any individual bit will be flipped is 0.1. What is the probability that
a message 4 bits long would be successfully transmitted? (Answer to three decimal places).
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One method of dealing with the problem of bits being flipped is to use a Hamming code. This involves sending extra bits along with the message that can
be used to check the main message. For example a 7 bit Hamming Code contains 4 bits of message data and 3 check bits. If only one of the bits is in
error at the receiving end then mathematical techniques can be used to determine which one it is and apply a correction. However, if more than one bit is
flipped then an erroneous correction will be applied and the message will still be corrupted.
Assume that a message is being sent through a noisy channel where the probability that any individual bit will be flipped is 0.1 as before. If the message is
sent using a 7 bit hamming code what is the probability that it will get through with no more then one of the seven bits being flipped? (Answer to three
decimal places).
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If we are concerned with the possibility that 2 bits have been flipped then instead of using the correction mentioned above we can ask for the 7 bits to be
re-transmitted. However, there is the possibility that when more then 3 bits have been flipped that the final corrupted message will correspond to a
message where the Hamming code thinks that none of the bits have been flipped. The probability of this occurring when 3 or 4 bits have been flipped is
0.2. It does not occur if 5 or 6 bits have been flipped (similar to 1 or 2 bits being flipped). It always occurs if all 7 of the bits have been flipped.
Assume that a message is being sent through a noisy channel where the probability that any individual bit will be flipped is 0.1 as before. A 7 bit Hamming
code is being used. When the message is decoded at the receiving end the procedure indicates that none of the bits have been flipped. What is the
probability that the message that was received that appears to have no bits flipped is in fact an error with 3 or more bits flipped? (Answer to three decimal
places).
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Assume that a message is being sent through a noisy channel where the probability that any individual bit will be flipped is 0.1 as before. A 7 bit Hamming
code is being used. When the message is decoded at the receiving end the procedure indicates that at least one of the bits have been flipped. Therefore
the 7 bit message is re-transmitted. What is the probability that if the decoding procedure is applied to both transmissions that both times will result in the
correct 7 bit message (including the case where the second transmission does not need any unflipping)? (Answer to three decimal places).
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Answer 1

Given probability of being flipped (i.e. changed from a 0 to a 1 or vice versa) resulting in a corrupted message is q = 0.1 (let it be treated as failure case)

Then probability of not getting a corrupted message is p = 0.9 ( success case)

{ The probability of success and the probability of failure sum to unity (one) }

Now using Bernoulli's trials, probability of k successes in n trials is given by

The probability of 4 bits not getting corrupted (k success cases) out of 4 bits is solved by substituiting n=4, k=4 , p=0.9 and q=0.1 in the above equation .

Therefore P(k) = * (0.9)⁴ *(0.1)^4-4 = 1 * (0.9)⁴ *1 = 0.6561

{ = 1 }

The probability that a message 4 bits long would be successfully transmitted = 0.656

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This problem is more likely the first one

Given number of bits(n) = 7

The probability of individual bit being flipped = p =0.1

The probability of individual bit not being flipped = q =0.9

Now the probability that it will get through with no more than one of the seven bits being flipped = probability of 0 bit being flipped + probability of 1 bit being flipped

Probability of flipping 0 bit (k=0) out of 7 bits (n=7) = * (0.1)⁰ * (0.9)^7-0 = 1*1*(0.9)⁷ = 0.478

Probability of flipping 1 bit (k=1) out of 7 bits (n=7) = * (0.1)¹* (0.9)^7-1 = 7*0.1*(0.9)⁶ = 0.372

Therefore the probability that it will get through with no more than one of the seven bits being flipped = 0.478 + 0.372 = 0.85

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