A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are drawn what is the probability that one is yellow and two are red?
Correct Answer: $$\frac{{12}}{{455}}$$
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When three marbles are drawn, the probability that one is yellow and two are red
$$\eqalign{ & = \frac{{\left( {{}^2{C_1}} \right)\left( {{}^4{C_2}} \right)}}{{{}^{15}{C_3}}} \cr & = \frac{{2 \times 4 \times 3 \times 3 \times 2}}{{1 \times 2 \times 15 \times 14 \times 13}} \cr & = \frac{{144}}{{5460}} \cr & = \frac{{12}}{{455}} \cr} $$
When three marbles are drawn, the probability that one is yellow and two are red
$$\eqalign{ & = \frac{{\left( {{}^2{C_1}} \right)\left( {{}^4{C_2}} \right)}}{{{}^{15}{C_3}}} \cr & = \frac{{2 \times 4 \times 3 \times 3 \times 2}}{{1 \times 2 \times 15 \times 14 \times 13}} \cr & = \frac{{144}}{{5460}} \cr & = \frac{{12}}{{455}} \cr} $$