Answer :
Answer: A.8488
Step-by-step explanation:
Poisson distribution formula :
[tex]P(X=x)=\dfrac{e^{-\mu}\mu^x}{x!}[/tex]
, where [tex]\mu[/tex] = mean of the distribution.
Let x be the random variable that represents the number of defects.
Given : The average number of defects per standard sheet of steel is 2.
For 3 sheets , the average number of defects per standard sheet of steel = 2 x 3 = 6
i.e. [tex]\mu=6[/tex]
Now , the probability that the first three units manufactured in this batch will contain at least a total of 4 defects will be :-
[tex]P(X\geq4)=1-P(X<4)\\\\= 1-P(X\leq3)\\\\ 1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))\\\\ =1-(\dfrac{e^{-6}6^0}{0!}+\dfrac{e^{-6}6^1}{1!}+\dfrac{e^{-6}6^2}{2!}+\dfrac {e^{-6}6^3}{3!})\\\\=1-(0.00247875+0.0148725+0.0446175+0.089235)\\\\=1-0.15120375\\\\=0.84879625\approx0.8488[/tex]
Hence, the probability that the first three units manufactured in this batch will contain at least a total of 4 defects will be 0.8488.
Thus , the correct option is A .8488.