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Errors caused by contamination on optical disks oc-
cur at the rate of one error every $$10^{5}$$ bits. Assume the errors
follow a Poisson distribution.
(a) What is the mean number of bits until five errors occur?
What is the standard deviation of the number of bits until
five errors occur?
(c) The error-correcting code might be ineffective if there are
three or more errors within $$10^{5}$$ bits. What is the probabil-
ity of this event?

- The rate = one error every $$10^{5}$$ bits
- errors have poisson distribution

$$\Rightarrow \lambda=\frac{1}{10^{5}}=10^{-5}$$

(a) $$r=5, \lambda=10^{-5}$$

$$\mu=\frac{r}{\lambda}=\frac{5}{10^{-5}}=5 \times 10^{5} \mathrm{bits}$$

(b) $$\sigma_{x}^{2}=\frac{r}{\lambda^{2}}=\frac{5}{10^{-10}} \Rightarrow \sigma_{x}=\sqrt{5 \times 10^{10}}$$

$$=223607 \ \mathrm{bits}$$

(c) $$y \rightarrow n$$: of errors within $$10^{5}$$ bit

$$y \rightarrow$$ poisson distribution

$$\lambda=1$$

$$P(y \geq 3)=1-P(y \leq 2)$$

$$=1-[P(y=0)+P(y=1)+P(y=2)]$$

$$f(x)=\frac{e^{-\lambda} \lambda^{y}}{y !}$$

$$=1-\left[\frac{1^{0} e^{-1}}{0 !}+\frac{e^{-1} 1^{1}}{1 !}+\frac{e^{-1} 1^{2}}{2 !}\right]$$

$$=0.0803$$

Use the properties of the gamma function to evaluate
the following:
(a) $$\Gamma(6) \quad$$ (b) $$\Gamma(5 / 2)$$
(c) $$\Gamma(9 / 2)$$

- evalute:

a) $$\Gamma(6)$$

b) $$\Gamma\left(\frac{5}{2}\right)$$

c) $$\Gamma\left(\frac{9}{2}\right)$$

$$\Gamma(n)=(n-1) !$$

$$\Gamma(n+1)=n\Gamma(n)$$

$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$

(a) $$\Gamma(6)=(6-1) !=5 !$$

$$=5 \times 4 \times 3 \times 2 \times 1=120$$

(b) $$\Gamma\left(\frac{5}{2}\right)=\frac{3}{2} \Gamma\left(\frac{3}{2}\right)=\frac{3}{2} \frac{1}{2} \Gamma\left(\frac{1}{2}\right)$$

$$=\frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} \simeq 1.329$$

(c) $$\Gamma\left(\frac{9}{2}\right)=\frac{7}{2} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \Gamma{(1 / 2)}$$

$$=\frac{105}{16} \sqrt{\pi}$$

$$\simeq11.632$$

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