, 2015

Solution:

The only information this gives us about the number of yellow balls left is that it is even. A bijection shows that the probability that there are $k$ yellow balls left is equal to the probability that there are $30-k$ yellow balls left (flip the colors of the red and blue balls, and then switch the 65 balls that have been picked with the 65 balls that have not been picked). So the expected number of yellow balls left is 15. Therefore the expected number of red balls left is 22.5. So the answer is $\frac{22.5}{65}=\frac{45}{130}=\frac{9}{26}$.

Solution:

Let $w(b)=\binom{50}{b}\binom{50}{r=b+5}\binom{30}{60-2b}$ be the number of possibilities in which $b$ blue balls have been drawn (precisely $15 \leq b \leq 30$ are possible). For fixed $b$, the probability of drawing red next is $\frac{50-r}{50+50+30-65}=\frac{45-b}{65}$. So we want to evaluate
$$\frac{\sum_{b=15}^{30} w(b) \frac{45-b}{65}}{\sum_{b=15}^{30} w(b)}$$

Combinatorics

Note the symmetry of weights:
$$w(45-b)=\binom{50}{45-b}\binom{50}{50-b}\binom{30}{2b-30}=\binom{50}{b+5}\binom{50}{b}\binom{30}{60-2b}$$
so the $\frac{45-b}{65}$ averages out with $\frac{45-(45-b)}{65}$ to give a final answer of $\frac{45 / 2}{65}=\frac{9}{26}$.