Answer:
The probability that the message will be wrong when decoded is 0.05792
Step-by-step explanation:
Consider the provided information.
To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.
We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.
The digit transmitted is incorrectly received with probability p = 0.2
The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8
We want the probability that the message will be wrong when decoded.
This can be written as:

Hence, the probability that the message will be wrong when decoded is 0.05792
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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1 liter = 0.1 decaliter
Divide the volume value by 10
N would equal 57 × 3 which is 171.