Just keep drawing

Let $A_j$ be the event that $j$ of the first $4$ picks are red (for $2 \le j \le 4$ ). Let $A$ be the event that at least $2$ of the first $4$ picks are red, and let $B$ be the event that the fifth pick is red. Certainly,then, $A = A_2 \cup A_3 \cup A_4$ . Thus $\begin{aligned} \mathbb{P}[A] & = \mathbb{P}[A_2] + \mathbb{P}[A_3] + \mathbb{P}[A_4] \; = \;\frac{\binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} + 1}{\binom{10}{4}} \; = \; \frac{115}{210} \\[2ex] \mathbb{P}[A \cap B] & = \mathbb{P}[A_2 \cap B] + \mathbb{P}[A_3 \cap B] + \mathbb{P}[A_4 \cap B] \\ & = \frac{\binom{4}{2}\binom{6}{2}}{\binom{10}{4}} \times \frac13 + \frac{\binom{4}{3}\binom{6}{1}}{\binom{10}{4}} \times \frac16 + \frac{1}{\binom{10}{4}} \times 0 \; = \; \frac{34}{210} \end{aligned}$ and hence $\mathbb{P}[B|A] \; = \; \frac{\mathbb{P}[A \cap B]}{\mathbb{P}[A]} \; = \; \frac{34}{115}$ making the answer $\boxed{149}$ .