(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability of obtaining :
(i) exactly 8 heads ; (ii) at least 5 heads ; (iii) at most 1 head.
(b) In how many ways can four letters from the word SHEEP be arranged (i) without any restriction ; (ii) with only one E.
(a) \(p(head) = p = \frac{1}{2}; p(tail) = q = \frac{1}{2}\)
\((p + q)^{8} = p^{8} + 8p^{7} q + 28p^{6} q^{2} + 56p^{5} q^{3} + 70p^{4} q^{4} + 56p^{3} q^{5} + 28p^{2} q^{6} + 8pq^{7} + q^{8}\)
(i) p(exactly 8 heads) = \(p^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)
(ii) p(5 heads) = \(56p^{5} q^{3} = 56(\frac{1}{2})^{5} (\frac{1}{2})^{3} = \frac{7}{32}\)
p(6 heads) = \(28p^{6} q^{2} = 28(\frac{1}{2})^{6} (\frac{1}{2})^{2} = \frac{7}{64}\)
p(7 heads) = \(8p^{7} q = 8(\frac{1}{2})^{7} (\frac{1}{2}) = \frac{1}{32}\)
p(at least 5 heads) = \(\frac{7}{32} + \frac{7}{64} + \frac{1}{32} + \frac{1}{256} = \frac{93}{256}\)
(iii) p(at most 1 head) = p(0 head) + p(1 head)
p(0 head) = \(q^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)
p(1 head) = \(8p q^{7} = 8(\frac{1}{2})(\frac{1}{2})^{7} = \frac{8}{256}\)
p(at most 1 head) = \(\frac{1}{256} + \frac{8}{256} = \frac{9}{256}\)
(b) The word SHEEP has 5 letters, two of which are identical.
(i) Hence 4 letters from the word SHEEP can be arranged in \(\frac{^{5} P_{4}}{2!}\) ways.
= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60\).
(ii) The two Es are identical. If one of them are taken, there are 4 letters to be arranged in \(^{4} P_{4}\) ways.
= \(\frac{4!}{(4 - 4)!} = 4! = 24\).
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