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7 May, 18:20

A bucket contains 5 green tennis balls, two yellow tennis balls, 6 red tennis balls, and 8 blue tennis balls. Tony removes 4 tennis balls, without replacement, from the bucket shown. What is the probability that Tony removes 1 yellow, 1 green, 1 blue, and then 1 more green tennis ball?

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Answers (2)
  1. 7 May, 19:13
    0
    Total amount of tennis balls:

    5 green tennis balls + 2 yellow tennis balls + 6 red tennis balls + 8 blue tennis balls = 21 tennis balls.

    The problem says that Tony removes 4 tennis balls without replacement, so:

    Let's call Event A = yellow, green, blue, green

    P (A) = (2/21) (5/20) (8/19) (4/18)

    P (A) = 2.23x10^-3

    P (A) = 0.00223

    P (A) = 0.223%
  2. 7 May, 21:21
    0
    Originally, we have

    5 green (G), 2 yellow (Y), 6 red (R), 8 blue (B) balls.

    Total number of balls = 5 + 2 + 6 + 8 = 21

    1st draw: 1 yellow drawn. the probability is

    P1 = 2/21

    We now have 5 G, 1 Y, 6 R, 8 B = 20 balls

    2nd draw: 1 green drawn. The probability is

    P2 = 5/20 = 1/4

    We now have 4 G, 1 Y, 6 R, 8 B = 19 balls.

    3rd draw: 1 blue drawn. The probability is

    P3 = 8/19

    We now have 4 G, 1 Y, 6 R, 7 B = 18 balls.

    4th draw: 1 green is drawn. The probability is

    P4 = 4/18 = 2/9

    The draws are independent events. Therefore the probability is

    P1*P2*P3*P4 = 0.00223

    Answer: 0.00223
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