(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).