Answer:
The probability that the message will be wrong when decoded is 0.05792
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:
![P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =(5!)/(3!2!)(0.2)^3(0.8)^(2)+(5!)/(4!1!)(0.2)^4(0.8)^(1)+(5!)/(5!)(0.2)^5(0.8)^0\\P(X\geq3) =0.05792](https://img.qammunity.org/2020/formulas/mathematics/college/6nainytlqdtsk2emv6r2io7ubfcjmnf4ym.png)
Hence, the probability that the message will be wrong when decoded is 0.05792