There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$ red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the $10$ ways to end the game, $4$ of them ends with a green chip. The answer is $\frac{4}{10}=\frac{2}{5}$, or $(\mathbf{B}) \frac{2}{5}$. 
A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn? 
  $$\frac{3}{10}$$ 
  $$\frac{2}{5}$$ 
  $$\frac{1}{2}$$ 
  $$\frac{3}{5}$$ 
  $$\frac{7}{10}$$ 