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.

  1. math_celebrity

    math_celebrity Administrator Staff Member

    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

     

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