Permutations probability

Question

There are 6 green marbles, 6 blue and 6 red. 5 marbles are picked at random without replacement.

1)What is the probability that the marbles are only one colour? 

2)what is the probability there's exactly two colours?

 

3) 6 marbles are now drawn. What is the probability that there are 3 different colours? 

Answer 1

The issue on problems like this is what is the easiest way to calculate the answers? We are helped in this problem by having the same number of each color.

1) P(only 1 color) = 3 P(only green) = 3 C(6,5)/C(18,5) (number of ways to choose 5 from green/number of ways to choose 5 from all colors)

3 C(6,5)/C(18,5) = 3 * 6/8568 = 1/476

2) P(having exactly 2 colors) = C(3,2)P(having green and blue)

P(having green and blue) = P(having green or blue) - P(having only green) - P(having only blue) = P(having green or blue) - 2 P(having only green) = (C(12,5) - 2 C(6,5))/C(18,5) = (792 - 2 * 6)/8568 = 780/7568 = 65/714

C(3,2)P(having green and blue) = 3* 65/714 = 65/238

Note that we can also multiply in the original numerator and have (3 C(12,5) - 6 C(6,5))/C(18,5)

3) We now shift to 6 marbles.

We can resolve 1 and 2, substituting 6 for 5, and subtract from 1.

Then, 1 - (3 C(6,6) + 3 C(12,6) - 6 C(6,6))/C(18,6) = 1 - (3 C(12,6) - 3 C(6,6))/C(18,6) =

1 - (2772 - 3)/18564 = 1 - 71/476 = 405/476

Just for grins, I thought it would be fun to find the numerator also by adding.

To have 6, with at least one of each, means that you can have 1, 1, 4 (3 ways); 1, 2, 3 (6 ways); or 2, 2, 2 (1 way);

Then, 3 C(6,1)C(6,1)C(6,4) + 6 C(6,1)C(6,2)C(6,3) + C(6,2)C(6,2)C(6,2) =

3 * 6^2 * 15 + 6 * 6 * 15 * 20 + 15^3 = 15795

15795/18564 = 405/476; answer is same!