There are 25 marbles: (12 red, 7 yellow, 5 blue, 1 white)
The probability of the first marble being red is 12/25 because there are 12 red marbles still available and no marbles are missing
The probability of the second marble being red is 11/24 because only 11 red marbles are available and 1 marble has already been selected from the pile
The probability of the third marble being red is 10/23 because only 10 red marbles are available and 2 marbles have already been selected from the pile
The probability of the fourth marble being red is 9/22 because only 9 red marbles are available and 3 marbles have already been selected from the pile
Since we are interested in all of these events occurring simultaneously, we must multiply the probability of all 4 events, like so:
[tex] \frac{12}{25} \times \frac{11}{24} \times \frac{10}{23} \times \frac{9}{22} [/tex]
Solving for this we are left with:
[tex] \frac{9}{230} \: \: or \: \: 0.0391[/tex]
