Abucket 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?

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%

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

0.00223

The answer would be (c)

Ithink the answer is b i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you