Finding probability related to Poisson and Exponential Distribution

[songoku]

Homework Statement

Please see below

Relevant Equations

Poisson Distribution

Exponential Distribution

My attempt:
(i) $\lambda =3$

(ii)
(a) $P(N_{2} \geq 1=1-P(N_{2} =0)=1-e^{-6} \frac{(-6)^0}{0!}=0.997$

(b) $P(N_{4} \geq 3)=1-P(N_{4} \leq 2)=0.999$

(c) $P(N_{1} \geq 2) = 1-P(N_{4} \leq 1)=0.8$

Do I even understand the question correctly for part (i) and (ii)?(iii) The expectation of exponential distribution is $\frac{1}{\lambda}$ so the answer is $\frac{1}{3}$?

Thanks

 

[mjc123]

Do I even understand the question correctly for part (i) and (ii)?

Yes, I think so, but there are a couple of errors in notation.
In (a), (-6)⁰ should be 6⁰.
In (b), N₄ should be N₁.
Otherwise it looks correct.
For (iii), what are you calculating the expectation of?

 

[FactChecker]

Here is a simple sanity check of your answer for (ii)(a):
On average, there are three failures per unit time. Your answer for part (ii)(a) is that it is almost certain that there are no failures in two time units. Does that seem right?

 

[mjc123]

D'oh! should have seen that!

 

[songoku]

Here is a simple sanity check of your answer for (ii)(a):
On average, there are three failures per unit time. Your answer for part (ii)(a) is that it is almost certain that there are no failures in two time units. Does that seem right?

Yeah, it does not make sense.

there are a couple of errors in notation.

In (b), N₄ should be N₁.

Sorry I don't understand why it should be N₁, maybe I am misunderstanding something again.

N = number of failures
t = time interval

Is that correct?Revised attempt:
(ii)
(a) $P(N_{2} = 0)=e^{-6}=0.0025$

(b) still the same as post #1

(c) still the same as post #1, with a revision $N_{4}$ should be $N_{1}$
Oh wait, maybe you mean the error in notation is for (c) instead of (b)? @mjc123

For (iii), what are you calculating the expectation of?

Oh my god, the expectation of Sn

$$E(S_{n}-S_{n-1})=\frac{1}{\lambda}$$
$$E(S_{n})-E(S_{n-1})=\frac{1}{\lambda}$$
$$E(S_{n})=\frac{1}{\lambda}+E(S_{n-1})$$

Not even sure this is the correct approach

Thanks

 

[mjc123]

Oh no, yes I meant c not b. I must have been half asleep that day. Apologies for confusion.
(iii), you're on the right track.
S₀ = 0
E(S₁ - S₀) = 1/λ
E(S₁) = 1/λ
E(S₂ - S₁) = 1/λ
E(S₂) = ?
etc.

 

[songoku]

I understand. Thank you very much for the help mjc123 and FactChecker