Question

A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:
(i) all 10 are defective
(ii) all 10 are good
(iii) at least one is defective
(iv) none is defective

Answer 1

Out of 100 bulbs, 10 can be chosen in ¹⁰⁰C₁₀ ways.
So, total number of elementary events = ¹⁰⁰C₁₀
(i)
There are 20 defective and 80 non-defective bulbs.
The number of ways of selecting 10 defective bulbs out of 20 is ²⁰C₁₀ ways.
∴ Favourable number of elementary events = ²⁰C₁₀ ways
Hence, required probability = $\frac{{}^{20}C_{10}}{{}^{100}C_{10}}$

(ii)
The number of ways of selecting 10 non-defective bulb out of 80 is ⁸⁰C₁₀ ways.
∴ Favourable number of elementary events = ⁸⁰C₁₀
Hence, required probability = $\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$

(iii)
Probability for at least one defective bulb = 1 – Probability (all 10 are non-defective)
= $1-\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$

(iv)
'None is defective' means that all are non-defective bulbs. The number of ways of selecting all 10 non-defective bulbs out
of 80 is ⁸⁰C₁₀ ways.
∴ Favourable number of elementary events = ⁸⁰C₁₀
Hence, required probability = $\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$