(a) $p(head)=p=\frac{1}{2};p(tail)=q=\frac{1}{2}$
$(p+q{)}^8=p^8+8p^{7}q+28p^6{q}^2+56p^5{q}^3+70p^4{q}^4+56p^3{q}^5+28p^2{q}^6+8p{q}^7+q^8$
(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.
= $\frac{4!}{(4-4)!}=4!=24$.