Question

a marble is picked at random. without putting the first marble back, a second marble is picked at random. find the probability that first marble picked is green and second marble picked is red. write your result as a percentage.

Explanation:

Step1: Calculate probability of first - green

There are 3 green marbles and 6 total marbles initially. So the probability of picking a green marble first, $P(G_1)=\frac{3}{6}$.

Step2: Calculate probability of second - red

After picking a green marble first (without replacement), there are 5 marbles left and 2 red marbles. So the probability of picking a red marble second given a green was picked first, $P(R_2|G_1)=\frac{2}{5}$.

Step3: Calculate joint probability

By the multiplication rule for dependent events $P(G_1\cap R_2)=P(G_1)\times P(R_2|G_1)=\frac{3}{6}\times\frac{2}{5}=\frac{6}{30} = 0.2$.

Step4: Convert to percentage

To convert 0.2 to a percentage, we multiply by 100. So $0.2\times100 = 20\%$.

Answer:

20%