Not sure about an exercise of probability (POISSON)

Question

I was given this exercise:

I did this:

I know that:

Fabric rolls of 10 meters² and according to POISSON 2 defects are expected for every 20 meters². Having 2 defects every 20 square meters means that on average there is 1 fault every 10 square meters.

A roll is of poor quality if it has 3 or more flaws. Therefore, the probability that the roll is of poor quality:

P(x >= 3) = 1 – P(x < 3)
              = 1 – P(x = 0) + P(x = 1) + P(x = 2)
          = 1 – e^(-1) – [(e^(-1).1)/(1)] - [(e^(-1).2)/(2!)]
          = 1 – e^(-1).(1 + 1 + ½)
          = 0.08025

Let ‘n’ be the variable that shows the number of rolls reviewed, of which 4 were found to be defective:

n * 0.08025 = 4.
n = (4 / 0.08025) = 49.84 and this is approximately 50 rolls.

What should i do from now to determine the acumulated amount of money?

Answer 1

Let $X$ be the earnings, then by definition of expected value,

$$\begin{split}\mathbb E(X)&=60 \mathbb P(X=60)+70\mathbb P(X=70) +80\mathbb P(X=80)\end{split}$$

Let $p=.0803$, the chance that there are $3$ or more defects on the roll.

• The inspector earns $\$60$ if he inspects 4, 5, or 6 total rolls, given by probability $p^4+4p^4(1-p)+{5\choose 2}p^4(1-p)^2$.

• The inspector earns \$70 if there are 7 rolls inspected, ${6 \choose 3}p^4(1-p)^3$.

• The inspector earns \$80 if he inspects more than 7 rolls.

So the expected earnings is $$60\left(p^4+4p^4(1-p)+{5\choose 2}p^4(1-p)^2\right) + 70\left({6 \choose 3}p^4(1-p)^3\right) + 80\left(1-\left(p^4+4p^4(1-p)+{5\choose 2}p^4(1-p)^2)+{6 \choose 3}p^4(1-p)^3\right)\right)=79.94757$$

p=.0803
60*(p^4+4*p^3*(1-p)+choose(5, 2)*p^4*(1-p)^2) + 70*choose(6, 3)*p^4*(1-p)^3+80*(1-(p^4+4*p^3*(1-p)+choose(5, 2)*p^4*(1-p)^2+choose(6, 3)*p^4*(1-p)^3))
79.94757