Question

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

A marble is taken at random from the bag, the colour is noted and then it is replaced.
Another marble is taken at random.
What is the probability of exactly 1 red?

please need ASAP

Answer 1

3 out of 12 chance i believe.

Answer 2

`(3)/(8)`

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
`(1)/(4)\cdot (9)/(12)=(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:

`(9)/(48)+(9)/(48)=(18)/(48)=\boxed{(3)/(8)}`