# Problem on probability given frame sizes and their error correction probabilities in a wireless system

I am stuck trying to solve the following question for a while

Bit error rate after demodulation in an wireless system = 1.0e-03.
The system has 4 possible frame sizes – 48 bytes, 96 bytes, 72 bytes and 36 bytes with probabilities = 0.4, 0.25, 0.2 and 0.15 respectively. An error correction scheme is implemented on frames with probabilities of correcting an erroneous frame = 0.6, 0.75, 0.7 and 0.5 respectively.
a) Find the average residual frame error rate.
b) Find the probability that, no frame has error after correction in a series of 25 frames.
c) If a frame is in error after correction, find the probability of its size being 48 bytes.

The main problem I am facing is understanding the role of the given bit error rate in this question.
I will lay down what I tried so far, I will start with
(c)

P(48F) = 0.4, P(96F) = 0.25, P(72F) = 0.2, P(36F) = 0.15
the probabilities of correcting an erroneous frame given the frame sizes are P(C|48F) = 0.6, P(C|96F) = 0.75, P(C|72F) = 0.7, P(C|36F) = 0.5

so the probabilities of not correcting an erroneous frame given frame size is 48 is
P(E|48F) = 1 - 0.6 = 0.4, same for others

probability of not correcting erroneous frame P(E) = P(E|48F)*p(48F) + P(E|96F)*p(96F) + P(E|72F)*p(72F) + P(E|36F)*p(36F)

using naive bayes,
p(48F|E) = P(E|48F)*p(48F)/P(E) = 0.448

This is where I am getting lost

b)

Probability of no frames having error
P(C) = 1 - P(E)
so P(no error in 25 frames) = P(C)^{25}

a)

average frame size = 48 * 0.4 + 96 * 0.25 + 72 * 0.2 + 36 * 0.15 = 63 bytes
so the average frame error rate should be = 63 * 0.001/8

This is what I tried, my problem is understanding how to use the bit error rate for finding the given probabilities