Chapter 6.3, Problem 29E

a)

To determine

To show that P(X=0) = $e^{-\lambda }$

Answer to Problem 29E

P(X=0) = $e^{-\lambda }$

Explanation of Solution

Given:

X be the event that occurs in a 1-second interval.

Calculation:

Let, Probability mass function of Poisson distribution is,

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

Therefore, just put $x$ =0 in above formula,

  $P(X=0)=e^{-\lambda }\frac{{\lambda }^0}{0!}=e^{-\lambda }$

b)

To determine

To explain why X=0 is same as T>1

Answer to Problem 29E

X=0 is same as T>1

Explanation of Solution

Given:

X be the event that occurs in a 1-second interval.

Calculation:

Here, X=0 mean there are no events occurred in 1 second. That is, the amount of time that elapses until the next event will be greater than 1.

That’s why X=0 is same as T>1.

c)

To determine

To show that P(X=0) is same as P(T>1)

Answer to Problem 29E

P(X=0)= P(T>1)

Explanation of Solution

Given:

X be the event that occurs in a 1-second interval.

Calculation:

Here, X=0 mean there are no events occurred in 1 second. That is, the amount of time that elapses until the next event will be greater than 1.

Therefore, P(X=0) = $e^{-\lambda }$

And P(T>1)= $e^{-\lambda }$

Both are same.

d)

To determine

To show that P(X=0) = $e^{-2\lambda }$ for 2-second interval.

Answer to Problem 29E

P(X=0) = $e^{-2\lambda }$

Explanation of Solution

Given:

X be the event that occurs in a 2-second interval.

Calculation:

Let, Probability mass function of Poisson distribution is,

  $P(X=x)=e^{-2\lambda }\frac{{(2\lambda )}^x}{x!}=e^{-2\lambda }$

Therefore, just put $x$ =0 in above formula,

  $P(X=0)=e^{-2\lambda }\frac{{(2\lambda )}^0}{0!}=e^{-2\lambda }$

e)

To determine

To explain why X=0 is same as T>2

Answer to Problem 29E

X=0 is same as T>2

Explanation of Solution

Given:

X be the event that occurs in a 2-second interval.

Calculation:

Here, X=0 mean there are no events occurred in 2 second. That is, the amount of time that elapses until the next event will be greater than 2.

That’s why X=0 is same as T>2.

f)

To determine

To show that P(X=0) = $e^{-t\lambda }$ for t-second interval.

Answer to Problem 29E

P(X=0) = $e^{-t\lambda }$

Explanation of Solution

Given:

X be the event that occurs in a t-second interval.

Calculation:

Let, Probability mass function of Poisson distribution is,

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

Therefore, just put x=0 in above formula,

  $P(X=0)=e^{-t\lambda }\frac{{(t\lambda )}^0}{0!}=e^{-t\lambda }$

g)

To determine

To explain why X=0 is same as T>t

Answer to Problem 29E

X=0 is same as T>t

Explanation of Solution

Given:

X be the event that occurs in a t-second interval.

Calculation:

Here, X=0 mean there are no events occurred in t second. That is, the amount of time that elapses until the next event will be greater than t.

That’s why X=0 is same as T>t.

h)

To determine

To show that P(X=0) is same as P(T>t)

Answer to Problem 29E

P(X=0) = P(T>1)

Explanation of Solution

Given:

X be the event that occurs in a t-second interval.

Calculation:

Here, X=0 mean there are no events occurred in 1 second. That is, the amount of time that elapses until the next event will be greater than 1.

Therefore, P(X=0) = $e^{-t\lambda }$

And P(T>t)= $e^{-t\lambda }$

Both are same.