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`        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`
one year ago

## Answers : (2)

```							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
```							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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