A box contains 6 balls of unknown colours. 3 balls are drawn and they are found to be white. Then the probability that all 6 balls are white balls is

Out of 6 balls 3 are drawn and found to be white. So, the box contains minimum 3 white balls. 

A=Out of 6 balls exactly 3 are white,

B=Out of 6 balls exactly 4 are white,

C=A=Out of 6 balls exactly 5 are white

D= all 6 balls are white

E=Selected  3 ball are white

$$\begin{gathered} P\left(D/E\right)=\frac{P\left(D\right)P\left(E/D\right)}{P\left(E\right)} \\ P\left(D/E\right)=\frac{P\left(D\right)P\left(E/D\right)}{P\left(A\right)P\left(E/A\right)+P\left(B\right)P\left(E/B\right)+P\left(C\right)P\left(E/C\right)+P\left(D\right)P\left(E/D\right)} \end{gathered}$$$$\begin{gathered} P\left(\mathrm{All} \mathrm{balls} \mathrm{are} \mathrm{white}/3 \mathrm{balls} \mathrm{are} \mathrm{white}\right)=\frac{{ }^6{C}_3}{{ }^3{C}_3{+}^4{C}_3{+}^5{C}_3{+}^6{C}_3} \\ \Rightarrow P\left(\mathrm{All} \mathrm{balls} \mathrm{are} \mathrm{white}/3 \mathrm{balls} \mathrm{are} \mathrm{white}\right)=\frac{20}{1+4+10+20}=\frac{20}{35}=\frac{4}{7} \end{gathered}$$Using Baye’s theorem Required probability $=\frac{4}{7}$