(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.

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(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal ...

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(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 (s i x) = p = \frac{1}{6}; p (not six) = q = \frac{5}{6}

(p + q)^{6} = p^{6} + 6 p^{5} q + 15 p^{4} q^{2} + 20 p^{3} q^{3} + 15 p^{2} q^{4} + 6 p q^{5} + q^{6}

(i) p(exactly 3 sixes) =

20 p^{3} q^{3}

20 (\frac{1}{6})^{3} (\frac{5}{6})^{3}

20 (\frac{1}{216}) (\frac{125}{216})

\frac{2500}{46656}

= 0.0536

≊

0.054.
(ii) p(at most 3 sixes) = p(0 six) + p(1 six) + p(2 sixes) + p(3 sixes)
=

q^{6} + 6 p q^{5} + 15 p^{2} q^{4} + 20 p^{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 ≊ 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 ≊ 0.960

(3 d.p)

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