A manufacturer, who produces medicine bottles, finds that 0.1% of the bottlesare defective. The bottles are packed in boxes containing 500 bottles. A drugmanufacturer buys 100 boxes from the product of bottles. Find how manyboxes will contain (i) no defective, and (ii) at least two defectives

Total bottles bought by manufacturer is equal to 50000 bottles.... 0.1 percent are defective. . . . . Therefore no. Of defective bottles is equal to 50 bottles.. . . . . Therefore probablity of boxes containing NO DEFECTIVE BOXES = 50/100 = 1/2.... . . Probablity of getting boxes containing defective bottles = 50/100 = 1/2.......

We know that:
P(X=x) = (μx e-μ)/x! for x = 0, 1, 2, …, ∞
Here;
e = 2.71828
n = 500
p = 0.1% or 0.001
μ = np
μ = 500 * 0.001
μ = 0.5
Our Poisson formula becomes:
p(x; 0.5) = (0.5x e-0.5)/x!
i. No defective
p(x; 0.5) = (0.5x e-50)/x!
p(0; 0.5) = (0.50 (2.71828)-0.5)/0!
p(0; 0.5) = (1 * 0.606531)
p(0; 0.5) = 0.606531 or 60.65% chances that there will be no defective.
ii. At least two defective
P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + ….
P(X ≥ 2) = 1 – P(X=1) – P(X=0)
So, first find P(X=1):
p(x; 0.5) = (0.5x e-0.5)/x!
p(1; 0.5) = (0.51 (2.71828)-0.5)/1!
p(1; 0.5) = (0.5 * 0.606531)/1
p(1; 0.5) = 0.303265 or 30.33% chances that there will be exactly one defective
P(X ≥ 2) = 1 – P(X=1) – P(X=0)
P(X ≥ 2) = 1 – 303265 – 606531
P(X ≥ 2) = 0.090204 or 9.02% chances that at least 2 will be defective