A jar contains 8 red, 6 blue, 4 green, and 2 yellow marbles. Assuming you don't replace the marbles, find the probability of pulling a green and then a green marble. Write your answer as a fraction in simplest form. Answer =

3 / 95

Step-by-step explanation:

Solution:-

- The jar contains colored marbles that are distributed according to colors as follows:

Color Number of marbles

Red 8

Blue 6

Green 4

Yellow 2

Total 20

- We are to draw 2 marbles from the jar. After each draw the colored marble are not replaced. The probability of drawing any colored marble in each trial is dependent on the previous draw.

- We will investigate each draw separately. The probability to draw a green marble in the first draw.

p (1: Green) = Favorable outcomes (Green) / Total number of marbles

- We have 4 Green marbles initially. So the number of favorable outcomes is 4.

The total number of balls in the jar are 20. Hence, the probability of drawing a green marble in first trial is:

p (1: Green) = 4 / 20 = 1 / 5

- Since, the drawn green marble is not replaced the color distribution also changes for the next draw of a colored ball. The new colored marble distribution is:

Color Number of marbles

Red 8

Blue 6

Green 3

Yellow 2

Total 19

- We see that we are left with 3 green marbles in the jar and the total is now 19. Therefore, the probability of drawing green marble in the second draw given that we have drawn a green marble in first draw is:

p (2: green / 1: Green) = Remaining green marbles / total marbles

p (2: green / 1: Green) = 3 / 19

- The combined probability of drawing a green marble in first trial and again drawing a green marble in the second trial can be determined from conditional probability of dependent events.

p (1: Green & 2: Green) = p (2: green / 1: Green) * p (1:Green)

p (1: Green & 2: Green) = (1/5) * (3/19)

p (1: Green & 2: Green) = 3 / 95 ... Answer