# A jar contains 7 red marbles, 8 green marbles, and 6 blue marbles. What is the probability that you

Discussion in 'Calculator Requests' started by math_celebrity, Oct 12, 2021.

A jar contains 7 red marbles, 8 green marbles, and 6 blue marbles. What is the probability that you draw 4 green marbles in a row if you do not replace the marbles after each draw?

The key phrase in this problem is do not replace.

Draw #1:
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 8
Total Marbles in the Jar = 7 red + 8 green + 6 blue = 21
P(Green) = 8/21

Draw #2:
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 8 - 1 = 7
Total Marbles in the Jar = 7 red + 7 green + 6 blue = 20
P(Green) = 7/20

Draw #3:
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 7 - 1 = 6
Total Marbles in the Jar = 7 red + 6 green + 6 blue = 19
P(Green) = 6/19

Draw #4:
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 6 - 1 = 5
Total Marbles in the Jar = 7 red + 5 green + 6 blue = 18
P(Green) = 5/18

We want P(Green, Green, Green, Green)
Because each draw is independent of all other draws, we multiply each draw to get the final probability
P(Green, Green, Green, Green) = P(Green on Draw 1) * P(Green on Draw 2) * P(Green on Draw 3) * P(Green on Draw 4) *
P(Green, Green, Green, Green) = 8/21 * 7/20 * 6/19 * 5/18
P(Green, Green, Green, Green) = 1680/143640

Using our fraction simplifier, we get:
P(Green, Green, Green, Green) = 2/171