Julia is about to draw a marble from a bag that contains red, white, a

1. The question asks us to find a working pair of number of added red marbles and the probability of picking one.

2. First of all, let the number of new red marbles be equal to n. Now let's write out the probability of choosing a red marble (P), \(P = \frac{# \ of \ red \ marbles}{ # \ of \ all \ marbles} = \frac{# \ of \ old \ red \ marbles + # \ of \ new \ red \ marbles}{(# \ of \ old \ red \ marbles + # \ of \ new \ red \ marbles) + # \ of \ white \ marbles + # \ of \ blue \ marbles} = \)
\( = \frac{10 + n}{(10 + n) + 20 + 30} = \frac{10 + n}{60 + n}\).

3. Since there is no other information, let's compare P to possible probabilities.

- \(P = 10\% \rightarrow \frac{10 + n}{60 + n} = 0.1 \rightarrow 10 + n = 6 + 0.1n \rightarrow 0.9n = -4 \rightarrow n \approx -4.4\), which doesn't work.
- \(P = 20\% \rightarrow \frac{10 + n}{60 + n} = 0.2 \rightarrow 10 + n = 12 + 0.2n \rightarrow 0.8n = 2 \rightarrow n = 2.5\), which doesn't work.
- \(P = 30\% \rightarrow \frac{10 + n}{60 + n} = 0.3 \rightarrow 10 + n = 18 + 0.3n \rightarrow 0.7n = 8 \rightarrow n \approx 11.4\), which doesn't work.
- \(P = 40\% \rightarrow \frac{10 + n}{60 + n} = 0.4 \rightarrow 10 + n = 24 + 0.4n \rightarrow 0.6n = 14 \rightarrow n \approx 23.3\), which doesn't work.
- \(P = 50\% \rightarrow \frac{10 + n}{60 + n} = 0.5 \rightarrow 10 + n = 30 + 0.5n \rightarrow 0.5n = 20 \rightarrow n = 40\), which works.

4. Our answer will be: Number of Marbles - 40 and Probability of Red (in %) - 50.