ABC26GN5355 · Probability

Clearly, $n(S)={ }^{52} C_{2}=\frac{(52 \times 51)}{2}=1326$. 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. $$\begin{array}{ll} \therefore & n\left(E_{1}\right)={ }^{26} C_{2}=\frac{(26 \times 25)}{(2 \times 1)}=325 ; n\left(E_{2}\right)={ }^{4} C_{2}=\frac{(4 \times 3)}{(2 \times 1)}=6 ; \\ & n\left(E_{1} \cap E_{2}\right)=2 C_{2}=1 \\ \therefore & P\left(E_{1}\right)=\frac{n\left(E_{1}\right)}{n(S)}=\frac{325}{1326} ; P\left(E_{2}\right)=\frac{n\left(E_{2}\right)}{n(S)}=\frac{6}{1326} ; \end{array}$$ $P\left(E_{1} \cup E_{2}\right)=\frac{1}{1326}$ $\therefore \quad P$ (both red or both kings) $=P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+$ $P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)$ $$=\left(\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}\right)=\frac{330}{1326}=\frac{55}{221} .$$