A box contains N coins, m of which is fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $\frac{1}{2}$, while it is $\frac{2}{3}$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. Then the probability that the coin drawn is fair, when first toss head, second toss tail is

Box contains m fair coins, N-m biased coins

F= Event of selecting a fair coin

B= Event of selecting a biased coin

$P\left(F\right)=\frac{m}{N},P\left(B\right)=\frac{N-m}{N}$ 
For a fair coin $\mathrm{P}(\mathrm{H}/\mathrm{F})=\mathrm{P}(\mathrm{T}/\mathrm{F})=\frac{1}{2}$for biased coin $\mathrm{P}(\mathrm{H}/\mathrm{B})=\frac{2}{3},\mathrm{P}(\mathrm{T}/\mathrm{B})=\frac{1}{3}$Let A₁ = event of drawing biased fair coin And A₂ = event of drawing biased coin
Let E = event of getting head on 1st toss, tails on 2^nd toss Using Baye’s Theorem $$\begin{gathered} \mathrm{P}\left(\frac{{\mathrm{A}}_1}{\mathrm{E}}\right)=\frac{{\displaystyle \frac{m}{N}}\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}}{\frac{m}{N}\times \left(\frac{1}{2}\times \frac{1}{2}\right)+{\displaystyle \frac{N-m}{N}}\left({\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{3}}\right)}= \\ \mathrm{P}\left(\frac{{\mathrm{A}}_1}{\mathrm{E}}\right)=\frac{9m}{9m+8(N-m)} \\ \mathrm{P}\left(\frac{{\mathrm{A}}_1}{\mathrm{E}}\right)=\left(\frac{9\mathrm{m}}{\mathrm{m}+8\mathrm{N}}\right) \end{gathered}$$