(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability...
(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.
Explanation
(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.
= \(\frac{4!}{(4 - 4)!} = 4! = 24\).