The probability $P_k$  that a family has k children is given by  $P_k=(1-2a)2^{-(k-1)},$ $k\ge 2$.  It is known that a family has two boys. Then which of the following is/are true?

Let $A_2$   denote 2 boys 
 $B_2$ denotes 2 girls 
$C_n$  denote, n children
 $P(C_2/A_2)=\frac{P\left(C_2\right)p(A_2/C_2)}{{\displaystyle \sum _{n=2}^{\infty }P\left(C_n\right)P(A_2/C_n)}}$    $P(B_2/A_2)=\frac{{\displaystyle \sum _{n=2}^{\infty }P\left(C_n\right)p(A_2\cap B_2/C_n)}}{{\displaystyle \sum _{n=2}^{\infty }P\left(n\right)P(A_2/C_n)}}$

$$\begin{gathered} P_n=P\left(C_n\right)=(1-2a){\left(\frac{1}{2}\right)}^{n-1},n=2,3,\mathrm{......} \\ P(A_2/C_n)={}^n{c}_2{\left(\frac{1}{2}\right)}^{n}P\left(A_2\right)=\sum _{n=2}^{\infty }P\left(C_n\right)P(A_2/C_n) \\ =(1-2a)\sum _{n=2}^{\infty }\left({}^n{c}_2\right){\left(\frac{1}{2}\right)}^{2n-1}=\frac{(1-2a)}{4^2}\sum _{n=2}^{\infty }n(n-1){\left(\frac{1}{4}\right)}^{n-2} \\ =\frac{(1-2a)}{4^2}\frac{2}{{(1-\frac{1}{4})}^3}=\frac{8(1-2a)}{27} P(A_2\cap B_2)=\sum _{n=4}^{\infty }P\left(c_n\right)p(A_2{B}_2/C_n) \\ =P\left(C_4\right)P(A_2{B}_2/C_4) (1-2a){\left(\frac{1}{2}\right)}^3\left({}^4{c}_2\right){\left(\frac{1}{2}\right)}^4=\frac{6(1-2a)}{128}=\frac{3(1-2a)}{64} \\ P(C_2/A_2)=\frac{\left(\frac{1-2a}{8}\right)}{\frac{8(1-2a)}{27}}=\frac{27}{64} P(B_2/A_2)=\frac{\frac{3(1-2a)}{64}}{\frac{8(1-2a)}{27}}=\frac{81}{512} \end{gathered}$$