Question 12
A box contains $5$ red pens, $6$ blue pens and $4$ green pens. Three pens are consecutively taken from the box, at random and without replacement. The probability that the first pen is red, the second pen blue and the third pen green, is
(A) $\frac{7}{225}$ (B) $\frac{1}{5}$ (C) $\frac{9}{82}$ (D) $\frac{4}{91}$ (E) $\frac{3}{14}$
(A) $\frac{7}{225}$ (B) $\frac{1}{5}$ (C) $\frac{9}{82}$ (D) $\frac{4}{91}$ (E) $\frac{3}{14}$