Theory question 106
Probability(a) A bag contains 10 red and 8 green identical balls. Two balls are drawn at random from the bag, one after the other, without replacement. Find the probablity that one is red and the other is green.
(b) There are 20% defective bulbs in a large box. If 12 bulbs are selected randomly from the box, calculate the probablity that between two and five are defective.
Show the model answer
Model answer
(a) RG or GR: 2 × (10/18)(8/17) = 160/306 = 80/153.
(b) Binomial n = 12, p = 0.2. 'Between two and five' (3 or 4): ¹²C₃(0.2)³(0.8)⁹ + ¹²C₄(0.2)⁴(0.8)⁸ = 0.2362 + 0.1329 ≈ 0.3691.
(If 2 and 5 are included: P(2) + … + P(5) ≈ 0.7057.)
Explanation
Two orders (red first or green first) both count.