(a)P(Bad Quality) = \(\frac{30}{10}\) = 0.3 and P(Good Quality) = 1 - 0.3 = 0.7
(a)(i) P(exactly 6) = \(^{12}C_6\) x (0.3)\(^6\) x (0.7)\(^6\) = 0.0792
(a)(ii) P(at least 4) = 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)]
Where P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = \(^{12}C_0\)(0.3)\(^6\)(0.7)\(^{12}\) + \(^{12}C_1\)(0.3)\(^1\)(0.7)\(^{11}\) + \(^{12}C_2\)(0.3)\(^2\)(0.7)\(^{10}\) + \(^{12}C_3\)(0.3)\(^3\)(0.7)\(^{9}\)
= 0.01384 + 0.071184 + 0.16779 + 0.2397
Therefore; P(at least 4) = 1 - 0.4925 = 0.5075
(a)(iii) P(no bad one) = \(^{12}C_0\)(0.3)\(^o\)(0.7)\(^{12}\) = 0.0138

(b) To find P(4 boys and 3 girls) = \(\frac{^8C_4 \times ^5C_3}{^{13}C_7}\)
= \(\frac{70 \times 10}{1716}\)
= \(\frac{175}{429}\)
= 0.4079