(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)