Question

There is a bag filled with 4 blue, 3 red and 5 green marbles.

Answer 1

Answer:

3 out of 12 chance i believe.

Answer 2

Answer:

$\frac{3}{8}$

Step-by-step explanation:

We'll need to use casework, since the problem stipulates there needs to be exactly one red drawn. Because we're drawing two marbles, we can either draw a red one first or a red one second. With casework, we'll add the probability a red marble is drawn the first draw and the probability a red marble is drawn the second draw.

There are $4+3+5=12$ marbles total. The probability of drawing a red marble is equal to the number of red marbles (3) divided by the total number of marbles (12). Therefore, the probability of getting a marble on the first draw is $3/12=1/4$. Since the marble is replaced, there are still 12 marbles, 4 blue, 3 red, and 5 green, for the second draw. We want to add the probability a red marble is drawn the first draw to the probability a red marble is drawn the second draw. Therefore, for the second draw, let's find the chances of not choosing a red marble. There is a $9/12$ chance that the marble chosen is not red. Therefore, the probably of this case happening is $\frac{1}{4}\cdot \frac{9}{12}=\frac{9}{48}$.

This case has the exact same probability as the other case, where a non-red marble is chosen the first draw and the red marble is now chosen the second draw ($9/12$ chance to choose a non-red marble and $1/4$ chance to choose a red marble).

Add these cases up:

$\frac{9}{48}+\frac{9}{48}=\frac{18}{48}=\boxed{\frac{3}{8}}$