Question

A box contains 12 bulbs of which "4" are defective. Three bulbs are drawn randomly. What is the probability that (i) All the "3" bulbs are defective? (ii) At least 2¹ of the bulbs chosen are defective? (iii) At most '2' of the bulbs chosen are defective? Answer fast with step by step explanation. Irrelevant answers will be reported. ​

Answer 1

Step-by-step explanation:

Total numbers of bulbs = n(s) = 12

No. of defective bulbs = n(a)

No. of good bulbs = n(b) = 12-4 = 8

P(a) = n(a)/n(s) = 4/12 = 1/3

1. P(All three bulbs are defective) = 1/3×1/3×1/3 =1/27

2. P(At least two bulbs are defective) = 1/3×1/3×2/3+1/3×1/3×1/3 2/27+1/27 = 3/27 = 1/9

3.P(At most 2 bulbs are defective) = 2/3×2/3×2/3+1/3×2/3×2/3+1/3×1/3×2/3

= 8/27+4/27+2/27 = 14/27