Question

In a certain communications system, there is an average of 1 transmission error per 10 seconds. Assume that the distribution of transmission errors is Poisson. The probability of 1 error in a period of one-half minute is approximately ________.

Answer 1

Answer:

The probability of 1 error in a period of 0ne - half minute is 0.1494

Step-by-step explanation:

Formula for poisson distribution:

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

If there is an average of 1 error in 10 seconds

In one-half minutes (i.e. 30 seconds), there will be an average of 30/10 errors = 3 errors

$\lambda = 3 errors\\x = 1$

$P (X = 1) = \frac{\exp^{-3} 3^{1} }{1!}$

1! = 1

$P (X = 1) = 3 \exp^{-3}$

P(X = 1) = 3 * 0.0498

P(X = 1) = 0.01494

Answer 2

Answer:

The probability of 1 error in a period of one-half minute is approximately 0.15 .

Step-by-step explanation:

We are given that in a certain communications system, there is an average of 1 transmission error per 10 seconds.

Let X = distribution of transmission errors

So, X ~ Poisson($\lambda$) , where $\lambda$ = average transmission error per 10 seconds = 1

i.e; X ~ Poisson($\lambda$ = 1)  

The Probability distribution of Poisson distribution is given by;

$P(X=x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; x = 0,1,2,3,....$

Since we have to find the probability for a period of one-half minute and we are given for a period of per 10 seconds.

Firstly, we need to convert $\lambda$ into period of one-half minute(30 seconds), i.e;

             $\lambda$ for per 10 seconds period = 1

             $\lambda$ for 1 second period = $\frac{1}{10}$

             $\lambda$ for 30 second period = $\frac{1}{10} \times 30$ = 3 errors

So, required X ~ Poisson($\lambda=3$)

Now, probability of 1 error in a period of one-half minute = P(X = 1)

  P(X = 1) = $\frac{e^{-3} \times 3^{1} }{1!}$ = $3 \times e^{-3}$ = 0.1494

Therefore, probability of 1 error in a period of one-half minute is approximately 0.15 or 15% .