A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red marble on the second draw from the bowl without replacing the marble from the first draw, is 

Consider, the events, $E_1,E_2$ 

  $E_1=$ red marbles in first draw

 $E_2=$  green marbles in first draw

A = red marble in second draw without replacement 

  $\begin{array}{r}P\left(E_1\right)=\frac{6}{9},P\left(E_2\right)=\frac{3}{9}\\ P\left(A/{\mathrm{E}}_1\right)=\frac{5}{8} ,P_1\left(A/{\mathrm{E}}_2\right)=\frac{6}{8}\end{array}$

  $\therefore$    Required probability

   $\begin{array}{r}P\left(A\right)=P\left(E_1\right)\times P\left(A/{\mathrm{E}}_1\right)+P\left(E_2\right)\times P\left(A/{\mathrm{E}}_2\right)\\ =\frac{6}{9}\times \frac{5}{8}+\frac{3}{9}\times \frac{6}{8}=\frac{30+18}{72}=\frac{48}{72}=\frac{2}{3}\end{array}$

  Hence, option (a) is correct .