Answer: Option D
Clearly,
n (S) = $$n{\text{ }}(S) = $$   $${}^{52}\mathop C\nolimits_2 = $$   $$\frac{{\left( {52 \times 51} \right)}}{2}$$   = 1326
Let $${{E_1}}$$ = event of getting both red cards
$${{E_2}}$$ = event of getting both kings
Then, $${{E_1}}$$ $$ \cap $$ $${{E_2}}$$ = event of getting 2 kings of red cards.
∴ $$n{\text{ }}({E_1}) = {}^{26}\mathop C\nolimits_2 = \frac{{\left( {26 \times 25} \right)}}{{\left( {2 \times 1} \right)}}$$     = 325 and
$$n{\text{ }}({E_2}) = {}^4\mathop C\nolimits_2 = \frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$     = 6
$$n\left( {{E_1} \cap {E_2}} \right) = {}^2{C_2} = 1$$
$$\therefore P({E_1}) = \frac{{n({E_1})}}{{n(S)}} = \frac{{325}}{{1326}}$$      and
$$P({E_2}) = \frac{{n({E_2})}}{{n(S)}} = \frac{6}{{1326}}$$
$$P({E_1} \cap {E_2}) = \frac{1}{{1326}}$$
∴ P (both red or both kings)
$$\eqalign{
& = P\left( {{E_1} \cup {E_2}} \right) \cr
& = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr
& = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr
& = \frac{{330}}{{1326}} \cr
& = \frac{{55}}{{221}} \cr} $$