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Grade: 12th pass
        
A manufacturer, who produces medicine bottles, finds that 0.1% of the bottles
are defective. The bottles are packed in boxes containing 500 bottles. A drug
manufacturer buys 100 boxes from the product of bottles. Find how many
boxes will contain (i) no defective, and (ii) at least two defectives
one year ago

Answers : (2)

Ankit
104 Points
							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.......
						
one year ago
Arun
21571 Points
							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
one year ago
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