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Probability of error per bit = $10^{-7}$

Probability of bit arriving correctly = $1-10^{-7}$

Probabiilty of a 64 B frame arriving correctly = $(1-10^{-7})^{64*8} = 0.9999488 $

Probability of a frame being damaged = $1 – 0.9999488$ = $5.12 * 10^{-5}$

Since data-rate is 11-Mbps, no. of frames transmitted per second = $(11 * 10^{6}) / (64 * 8) = 21484.375$ frames/sec

No. of damaged frames per second = no. of frames/sec * prob. of a frame being damaged
 $= 21484.375 * (5.12 * 10^{-5}) = 1.0999 = 1.1$ frames/sec (approx.)
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I guess the approach will be:

$1 Mb=2^{20} bits \approx 10^{6} bits\ (This\ will\ make\ our\ calculations\ easier)\\ 64\ byte=64*8\ bits\\ No. \ of \ frames\ that\ can \ be\ transmitted\ per\ sec(fps)=11*10^{6} / (64 * 8)\\ Frame\ error\ rate(fer)=10^{-7} * 64 * 8 \\ \therefore average\ error= fps*fer=10^{-7} * 11 * 10^{6}=1.1\approx 1\ frame\ will\ be\ damaged\ per\ sec$

Even if you take 1Mb=2^20 bits,the ans will come as 1.15 which is approximately 1.

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