This is a problem related to error probability in a discrete communication system using a binary symmetric channel (BSC). It provides the following:

• A source with six different symbols (A, B, C, D, E, F), each assigned a binary code.
• A BSC with a bit error probability of $p = 0.01$, where $p$ represents the probability of a bit error during transmission.

We are asked to calculate the probabilities of certain transmission outcomes.

Solution Overview:

1. Code Assignment:

   • A: 0101
   • B: 0100
   • C: 0010
   • D: 0111
   • E: 0000
   • F: 1111

2. Transmission Errors:

   • The channel introduces errors with a probability of $p = 0.01$ per bit (each bit can either be transmitted correctly or flipped).
   • For each bit in the codeword, the probability of no error is $1 - p = 0.99$, and the probability of an error (bit flip) is $p = 0.01$.

Now let's address the questions:

a) Probability that all four symbols are transmitted without errors.

Each symbol has 4 bits, so the probability that each bit is transmitted correctly is $0.99$. For four bits (i.e., four symbols), the total probability for no errors is:

$P(\text{no error for all bits}) = (0.99)^4$

b) Probability that each of the four symbols has at least one bit transmitted correctly.

For this, we need to calculate the probability that a symbol has no bit error. The probability that a symbol has at least one bit transmitted correctly is $1 - P(\text{all bits wrong})$.

The probability of all bits being wrong for a symbol is $(0.01)^4$. Therefore:

$P(\text{at least one correct bit}) = 1 - (0.01)^4$

c) Probability that two symbols are transmitted without errors, and two symbols with at least one bit error.

This involves two cases:

• Two symbols are transmitted without error.
• Two symbols have at least one bit error.

We calculate the probability of each case and combine them.

Would you like me to proceed with the detailed calculations for these probabilities?