Answer:
[tex]P(x\geq 3)=0.3233[/tex]
Step-by-step explanation:
If the number of defects in poured metal follows a Poisson distribution, the probability that x defects occurs is:
[tex]P(x)=\frac{e^{-m}*(m)^{x}}{x!}[/tex]
Where x is bigger or equal to zero and m is the average. So replacing m by 2, we get that the probability is equal to:
[tex]P(x)=\frac{e^{-2}*(2)^{x}}{x!}[/tex]
Finally, the probability that there will be at least three defects in a randomly selected cubic millimeter of this metal is equal to:
[tex]P(x\geq 3)=1-p(x\leq 2)\\[/tex]
Where [tex]P(x\leq 2)=P(0)+P(1)+P(2)[/tex]
So, P(0), P(1) and P(2) are equal to:
[tex]P(0)=\frac{e^{-2}*(2)^{0}}{0!}=0.1353\\P(1)=\frac{e^{-2}*(2)^{1}}{1!}=0.2707\\P(2)=\frac{e^{-2}*(2)^{2}}{2!}=0.2707[/tex]
Finally, [tex]P(x\leq2)[/tex] and [tex]P(x\geq3)[/tex] are equal to:
[tex]P(x\leq 2)=0.1353+0.2707+0.2707=0.6767\\P(x\geq 3)=1-0.6767=0.3233[/tex]
