On a public switched telephone network using a 4800 bps modem, the bit error probability is 10^-3. Without any error detection, what is the approximate residual frame error rate for 9-bit frames (i.e., probability that a frame is received with one or more bit errors)?

Introduction / Context:
This problem checks understanding of how independent bit errors aggregate into a frame error rate when no error detection or correction is present. The key idea is that a frame is in error if any bit within that frame is flipped during transmission.

Given Data / Assumptions:

• Bit error probability p = 10^-3 on a typical public telephone line with a 4800 bps modem.
• Frame size = 9 bits.
• Bit errors are independent and identically distributed.
• No error detection; any corrupted frame goes undetected (residual frame error = probability of at least one bit error).

Concept / Approach:

The probability that a frame is received correctly equals the probability that all bits are correct. For independent bits, that is (1 - p)^n, where n is the number of bits in the frame. The frame error probability is then 1 - (1 - p)^n.

Step-by-Step Solution:

Verification / Alternative check:

Using the small-p approximation for at least one error: 1 - (1 - p)^n ≈ n * p for n * p ≪ 1. Here, n * p = 9 * 0.001 = 0.009, which matches the exact computation closely, validating the result.

Why Other Options Are Wrong:

0.003: Corresponds to about 3 errors per 1000 bits, not per 9-bit frame.

0.991 and 0.999: These are probabilities of a correct frame (approximate) or of no error, not residual error.

None of the above: Incorrect since 0.009 matches the correct evaluation.

Common Pitfalls:

Confusing bit error rate with frame error rate, and forgetting that any single bit flip corrupts the entire frame when no detection is present.

Final Answer:

0.009