Question

Suppose that minor errors occur on a computer in a space station, which will require re-calculation. Assume the occurrence of errors follows a Poisson process with a rate of 1/2 per hour. (a) Find the probability that no errors occur during a day. (b) Suppose that the system cannot correct more than 25 minor errors in a day, in which case a critical error will arise. What is the probability that a critical error occurs since the start of a day? Keep up to the 6th decimal place in your answer. (c) Suppose the error correction protocols reset themselves so long as there are no more than five minor errors occurring within a 2 hour window. The system just started up and an error occurred. What is the probability the next reset will occur within 2 hours?

Answer 1

Answer:

a

  $P(X = 0) = 0.6065$

b

$P(x < 25 ) = 1.18 *10^{-33}$

c

 $P(x \le 5 ) = 0.9994$    

Step-by-step explanation:

From the question we are told that

  The rate  is $\lambda = \frac{1}{2}\ hr^{-1}$    =  0.5 / hr

  Generally  Poisson distribution formula is mathematically represented as

       $P(X = x) = \frac{(\lambda t) ^x e^{-\lambda t }}{x!}$

Generally the probability that no error occurred during a day is mathematically represented as  

Here  t =  1  hour according to question a

So

   $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$

Hence

   $[tex]P(X = 0) = \frac{\frac{1}{2} ^0 e^{-\frac{1}{2}}}{0!}$

=>  $P(X = 0) = 0.6065$

Generally the probability that  a critical error occurs since the start of a day is mathematically represented as

Here  t =  1  hour according to question a

So

   $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$

Hence

      $P(x \ge 25 ) = 1 - P(x < 25 )$

Here

     $P(x < 25 ) = \sum_{x=0}^{24} \frac{e^{-\lambda} * \lambda^{x}}{x!}$

=>   $P(x < 25 ) = \frac{e^{-0.5} *0.5^{0}}{0!} + \cdots + \frac{e^{-0.5} *0.5^{24}}{24!}$

$P(x < 25 ) = 0.6065 + \cdots + \frac{e^{-0.5} *0.5^{24}}{6.204484 * 10^{23}}$

$P(x < 25 ) = 0.6065 + \cdots + 6.0*10^{-32}$

$P(x < 25 ) = 1.18 *10^{-33}$

Considering question c

Here  t =  2  

Gnerally given that the  system just started up and an error occurred  the probability the next reset will occur within 2 hours

$P(x \le 5 ) = \sum_{n=0}^{5} \frac{(\lambda t) ^x e^{-\lambda t }}{x!}$

=> $P(x \le 5 ) = \frac{(0.5 * 2) ^ 0 e^{- 0.5 * 2 }}{0!} + \cdots + \frac{(0.5 * 2) ^ 5 e^{- 0.5 * 2 }}{5!}$

=> $P(x \le 5 ) = \frac{1* 2.7183 }{1 } + \cdots + \frac{1 *2.7183 }{120}$

=>  $P(x \le 5 ) = 2.7183 + \cdots + 0.0226525$

 $P(x \le 5 ) = 0.9994$