a) ${H_0}:P\left( 6 \right) = \frac{1}{6}$

${H_1}:P\left( 6 \right) \gt \frac{1}{6}$

b) ${\left( {\frac{5}{6}} \right)^{10}} + 10 \times {\left( {\frac{5}{6}} \right)^9} \times \frac{1}{6} + \left( \begin{gathered}  10 \hfill \\  \,\,2 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{5}{6}} \right)^8} \times {\frac{1}{6}^2} + \left( \begin{gathered}  10 \hfill \\  \,3 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{5}{6}} \right)^7} \times {\left( {\frac{1}{6}} \right)^3}$

$1 - ({\left( {\frac{5}{6}} \right)^{10}} + 10 \times {\left( {\frac{5}{6}} \right)^9} \times \frac{1}{6} + \left( \begin{gathered}  10 \hfill \\  \,2 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{5}{6}} \right)^8} \times {\left( {\frac{1}{2}} \right)^2}$

$ + \left( \begin{gathered}  10 \hfill \\  \,3 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{5}{6}} \right)^7} \times {\left( {\frac{1}{6}} \right)^3})$

$ = 0.0697$ (3 sfs)

c) Die biased towards a six but result $ \lt 4$ so no evidence of bias

d) P(0, 1, 2 or 3 sixes)

$({\left( {\frac{1}{2}} \right)^{10}} + 10 \times {\left( {\frac{1}{2}} \right)^9} \times \frac{1}{2} + \left( \begin{gathered}  10 \hfill \\  \,\,2 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{1}{2}} \right)^8} \times {\left( {\frac{1}{2}} \right)^2} + \left( \begin{gathered}  10 \hfill \\  \,3 \hfill \\ \end{gathered}  \right) \times {\left( {\frac{1}{2}} \right)^7} \times {\left( {\frac{1}{2}} \right)^3})$

$ = 0.172$ or $11/64$