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:
[tex]P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792[/tex]
Hence, the probability that the message will be wrong when decoded is 0.05792
