(a) A fair die with six faces is thrown six times. Calculate, correct to three...
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining :
(i) exactly three sixes ; (ii) at most three sixes.
(b) Eight percent of screws produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that :
(i) exactly two will be defective ; (ii) not more than two will be defective.
Explanation
(a) \(p(six) = p = \frac{1}{6} ; p(\text{not six}) = q = \frac{5}{6}\)
\((p + q)^{6} = p^{6} + 6p^{5}q + 15p^{4}q^{2} + 20p^{3}q^{3} + 15p^{2}q^{4} + 6pq^{5} + q^{6}\)
(i) p(exactly 3 sixes) = \(20p^{3}q^{3}\)
= \(20(\frac{1}{6})^{3} (\frac{5}{6})^{3}\)
= \(20(\frac{1}{216})(\frac{125}{216})\)
= \(\frac{2500}{46656}\)
= 0.0536 \(\approxeq\) 0.054.
(ii) p(at most 3 sixes) = p(0 six) + p(1 six) + p(2 sixes) + p(3 sixes)
= \(q^{6} + 6pq^{5} + 15p^{2}q^{4} + 20p^{3}q^{3}\)
= \((\frac{5}{6})^{6} + 6(\frac{1}{6})(\frac{5}{6})^{5} + 15(\frac{1}{6})^{2} (\frac{5}{6})^{4} + 20(\frac{1}{6})^{3} (\frac{5}{6})^{3}\)
= \(\frac{46250}{46656}\)
= \(0.991\) (to 3 d.p)
(b) p(a defective screw) = 0.08 ; p(non- defective screw) = 1 - 0.08 = 0.92.
The binomial distribution = \((p + q)^{10}\).
(i) p(exactly 2 defective) = \(^{10}C_{2} (0.08)^{2} (0.92)^{8}\)
= \(45(0.0064)(0.5132)\)
= \(0.1478 \approxeq 0.148\)
(ii) p(not more than 2 defective) = p(0 defective or 1 defective or 2 defective)
= \(^{10}C_{0} p^{0} q^{10} + ^{10}C_{1} p^{1} q^{9} + ^{10}C_{2} p^{2} q^{8}\)
= \(1(0.92)^{10} + 10(0.08)(0.92)^{9} + 0.1478\)
= \(0.4344 + 0.3777 + 0.1478\)
= \(0.9599 \approxeq 0.960\) (3 d.p)