probability of detecting burst error of size 15 = $1 - (\frac{1}{2})^r = 1 - (\frac{1}{2})^8 $

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+1 vote

I think for (C), it should be The probability of "not detecting" a burst error of size 9 is $\frac{1}{2^7}$

And for (D), the probability of detecting burst error of size 15 should be $1-\frac{1}{2^8}$

Correct me if I am wrong.

+2 votes

**IF L is length of burst error & polynomial generator has highest degree as r.**

**1.All burst errors with L ≤ r will be detected.**

**2.All burst errors with L = r + 1 will be detected with probability 1 – (1/2) ^{r–1}.**

**3.All burst errors with L > r + 1 will be detected with probability 1 – (1/2) ^{r}.**

Now , as r = 8 here so it will be able to detect burst error of size 6 and it will dtect single bit error so option a and b are true

Now probabiltiy of detecting burst error of size 9 = 1-(1/2)^8 -------[by second rule as L = r+1, L=9 and r = 8]

and probabiltiy of detecting burst error of size 15 = 1-(1/2)^15-----[by third rule as L>r+1, L=15 and r = 8]

so both option c and option d are false

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