(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 probability 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 probability that between two and five are defective.
Explanation
No. of red = 10, No of green = 8 Total = 18 P(RG) = P(R) x P(G) = \(\frac{10}{18} \times \frac{8}{17} = \frac{40}{152}\) P(GR) = P(G) x P(R) = \(\frac{8}{18} \times \frac{10}{17} = \frac{40}{153}\) P(one red and one green) = \(\frac{40}{153} + \frac{40}{153} = \frac{80}{153}\)
(b) P = 20% = \(\frac{20}{100}\) = 0.2 q = 1 - P = 1 - 0.2 = 0.8 n = 12 P(2 \(\leq x \leq 5\)) = P(5) - P(2) = \(^{12}C_5(0.5)^5 (0.8)^7 - ^{12}C_2(0.2)^2(0.8)^{10}\) = 792 x (0.2)\(^5\) (0.8)\(^7\) - 65(0.2)\(^2\)(0.8)\(^10\) = 0.04892 - 0.2791 = 0.2302
