During an assembly process, parts arrive just as they are needed. However at one station, the probability is .01
that a defective part will arrive in a one-hour period. Find the probability that

(a) exactly 1 defective part arrives in a 4-hour span.

(b) 1 or more defective parts arrive in a 4-hour span

(c) exactly 1 defective part arrives in a 4-hour span and exactly 1 defective part arrives in the next 4 spans.

Question

During an assembly process, parts arrive just as they are needed. However at one station, the probability is .01
that a defective part will arrive in a one-hour period. Find the probability that

(a) exactly 1 defective part arrives in a 4-hour span. 

(b) 1 or more defective parts arrive in a 4-hour span

(c) exactly 1 defective part arrives in a 4-hour span and exactly 1 defective part arrives in the next 4 spans.

anonymous · Answer

I think I have to use poisson process. 

$$P(x,lamda)=e^{-lamda}lamda^{x}/x!$$

anonymous · Answer

The average for 1 defective part in a one-hour period is $$\mu=.01$$, so for 1defective within a 4-hour time span, it would be $$mu*T=lamda$$ or $$.01*4=.04$$

anonymous · Answer

from here I have to plug in lamda and x (which is 1) to the poisson process formula: 

$$f(1;.04)=.04^1*e^{-.04}/1!=.0384$$

anonymous · Answer

Is this right? My book got .0388 or .039 if you round it.

anonymous · Answer

If this is true than for the second part would I have to use:

f(1;.04)+f(2;.04) to figure out (b)? And if so how many time do I have to add, just by f(x+1,lamda), or more?