In the daily production of a certain kind of metal beam, the number of defects per foot on each beam, Y, follows a Poisson distribution with mean 0.5. The profit per foot on a beam when it is sold is given by X, where X = 30 – 2Y – $3Y ^2$. In the daily production of a certain kind of metal beam, the number of defects per foot on each beam, Y, follows a Poisson distribution with mean 0.5. The profit per foot on a beam when it is sold is given by X, where X = 30 – 2Y – $3Y^2$.
Suppose a bunch of 2-foot metal beams are to be inspected one at a time. What is the probability the inspector will find a perfect beam before one with at least two defects?
I'm not sure how to approach this? I know lambda would be equal to 1 instead of 0.5 and I know P(X=0) = 0.3679 but I don't understand how to proceed.
$P(X=0) = e^{-1} \approx 0.3679$ is indeed correct
$P(X=1) = e^{-1} \approx 0.3679$ is also true, so $P(X\ge 2) = 1-2e^{-1} \approx 0.2642$
The probability the inspector will find a perfect beam before one with at least two defects is then $\dfrac{P(X=0)}{P(X=0)+P(X\ge 2)} = \dfrac{ e^{-1}}{ e^{-1}+ 1-2 e^{-1}} = \dfrac{1}{e-1} \approx 0.5820$, since either one of those cases is found or the inspector looks at the next beam