(a) Two Mathematics books, 5 different Physics books and 3 different Chemistry are to be
(a) Two Mathematics books, 5 different Physics books and 3 different Chemistry are to be arranged on a shelf. How many arrangements are possible if ;
(i) books on the same subject must stand together? (ii) only the Physics books must stand together?
(b) In a certain community, 13 out of every 20 persons speak English. If 8 persons are selected at random from the community, find, correct to three significant figures, the probability that at least 3 of them speak English.
Explanation
(a) 2M, 5P, 3C.
(i) Tie same books together
Maths books can be arranged within themselves in 2! ways, Physics books can be arranged in 5! ways, Chemistry books in 3! ways.
The three sets of subjects can be arranged in 3! ways.
\(\therefore\) Total arrangements = 3!2!5!3!
= \(6 \times 2 \times 120 \times 6 = \text{8640 ways}\).
(ii) Tie Physics books together.
There are 1 + 2 + 3 = 6 books to be arranged in 6! ways.
The Physics books can be arranged in 5! ways.
\(\therefore\) Total arrangement = \(6!5! = 720 \times 120 = \text{86,400 ways}\)
(b) p(English speaking) = p = \(\frac{13}{20}\); p(non- English speaking) = q = \(\frac{7}{20}\)
8 persons selected,
\((p + q)^{8} = p^{8} + 8p^{7}q + 28p^{6}q^{2} + 56p^{5}q^{3} + 70p^{4}q^{4} + 56p^{3}q^{5} + 28p^{2}q^{6} + 8pq^{7} + q^{8}\)
p(at least 3 speak English) = 1 - [none + one + two].
= \(1 - [q^{8} + 8pq^{7} + 28p^{2}q^{6}]\)
= \(1 - [(0.35)^{8} + 8(0.65)(0.35)^{2} + 28(0.65)^{2}(0.35)^{2}]\)
= \(1 - [0.0002252 + 0.003346 + 0.1131]\)
= \(1 - 0.001488 = 0.98512\)
\(\approxeq 0.985\) (3 sig. figs).