given: box with 100 bulbs of which, 20 are defective

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

ten bulbs are drawn at random for inspection, therefore 

total possible outcomes are ¹⁰⁰C₁₀

therefore n(S)= ¹⁰⁰C₁₀ 

(i) let E be the event that all ten bulbs are defective

n(E)= ²⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{20_{c_{10}}}{100_{c_{10}}}\)

(ii) let E be the event that all ten good bulbs are selected

n(E)= ⁸⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)

(iii) let E be the event that at least one bulb is defective 

E = {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs 

Let E’ be the event that none of the bulb is defective

n(E')= ⁸⁰C₁₀

P(E') = \(\frac{n(E')}{n(S)}\)

P(E') = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)

Therefore, 

P(E) = 1-P(E’)

P(E) = 1-P(E’)

P(E) = \(1-\frac{80_{c_{10}}}{100_{c_{10}}}\)

(iv) let E be the event that none of the selected bulb is defective

n(E)= ⁸⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)