A box contains 15 red, 20 blue, and 16 green jelly beans. Jack first chooses a green jelly bean and eats it. Then he chooses a blue jelly bean and eats it. If each of the remaining jelly beans is equally likely to be chosen, what is the probability that Jack chooses a red jelly bean next?
(A) $\frac{15}{31}$
(B) $\frac{34}{49}$
(C) $\frac{15}{49}$
(D) $\frac{2}{7}$
(E) $\frac{1}{15}$
Before Jack eats any jelly beans, the box contains $15+20+16=51$ jelly beans.

After Jack eats 2 jelly beans, there are $51-2=49$ jelly beans remaining in the box.

One of the jelly beans that Jack ate was green, and the other was blue.

Thus after eating the 2 jelly beans, there are still 15 red jelly beans in the box.

If each of the remaining jelly beans is equally likely to be chosen, the probability that Jack chooses a red jelly bean next is $\frac{15}{49}$.

ANSWER: $(\mathrm{C})$

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