Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are king, is-

Answer: Option D

Clearly,
n (S) = $$n(S)=$$   $${}^{52}{C}_2=$$   $$\frac{\left(52\times 51\right)}{2}$$   = 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.
∴ $$n(E_1)={}^{26}{C}_2=\frac{\left(26\times 25\right)}{\left(2\times 1\right)}$$     = 325 and
$$n(E_2)={}^4{C}_2=\frac{\left(4\times 3\right)}{\left(2\times 1\right)}$$     = 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)
$$\begin{array}{rl}& =P\left(E_1\cup E_2\right)\\ & =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1\cap E_2\right)\\ & =\left(\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}\right)\\ & =\frac{330}{1326}\\ & =\frac{55}{221}\end{array}$$