A. Four fair dice \( D _ { 1 } , D _ { 2 } , D _ { 3 } \) and \( b \)

Four fair dice $D_1,D_2,D_3$ and $D_4$, each having six faces numbered $1,2,3,4,5$ and $6,$ are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1,D_2$ and $D_3$ is

Four fair dice $D_1,D_2,D_{3}and{D}_4$ each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1,D_{2}and{D}_3$, is ?

Here are three vectors in meters:
$$ \overrightarrow {d_1} =- 3.0 \hat {i} + 3.0 \hat {j} + 2.0 \hat {k} $$
$$ \overrightarrow {d_2} =- 2.0 \hat {i} - 4.0 \hat {j} + 2.0 \hat {k} $$
$$ \overrightarrow {d_3}= 2.0 \hat {i} + 3.0 \hat {j} + 1.0 \hat {k} $$
What results from (a) $$ \overrightarrow {d_1}. ( \overrightarrow {d_2} + \overrightarrow {d_3} ) , (b) \overrightarrow {d_1} . ( \overrightarrow {d_2 } \times \overrightarrow {d_3}) $$ and (c) $$\overrightarrow {d}_1$$$$ \times (\overrightarrow {d_2} + \overrightarrow {d_3} ) $$?

If $$ \overrightarrow{d_1}+\overrightarrow{d_2}=5\overrightarrow{d_3},\overrightarrow{d_1}-\overrightarrow{d_2}=3\overrightarrow{d_3}, $$ and $$ \overrightarrow{d_3}= 2 \hat{i}+ 4\hat{j} $$ then what are, in unit-vector notation,(a) $$ \overrightarrow{d_1} $$ and (b) $$ \overrightarrow{d_2} $$?

Let $$D = diag [d_{1}, d_{2}, d_{3}]$$, where none of $$d_{1}, d_{2}, d_{3}$$ is $$0$$, prove that $$D^{-1} = diag [d_{1}^{-1}, d_{2}^{-1}, d_{3}^{-1}]$$.