ABC26GN5335 · Probability

We have $n(s)={ }^{52} C_{2}=\frac{(52 \times 51)}{(2 \times 1)}=1326$. Let $A=$ event of getting both black cards; $B=$ event of getting both queens. $\therefore A \cap B=$ event of getting queens of black cards. $\therefore n(A)={ }^{26} C_{2}=\frac{(26 \times 25)}{(2 \times 1)}=325, n(B)={ }^{4} C_{2}=\frac{(4 \times 3)}{(2 \times 1)}=6$ and $n(A \cap B)={ }^{2} C_{2}=1$. $\therefore P(A)=\frac{n(A)}{n(S)}=\frac{325}{1326} ; P(B)=\frac{n(B)}{n(S)}=\frac{6}{1326}$ and $P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{1326}$. $\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B)=\left(\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}\right)=\frac{330}{1326}=\frac{55}{221}$.