A bag contains 12 marbles, 7 red and 5 green. The reds are numbered 1-7 and the green are numbered 1-5. If you pick 4 without replacing what is the probability that 2 of the 4 are numbered 5?

I think it is 1/11 but don't know how

1/11 is correct.

We can do it in two different ways, easy and difficult.

Easy way:
Pick 2 of the 5-marbles and any of the remainder.
2/12*1/11=2/132=1/66.
The two 5-marbles can be arranged in 4P2 = 4!/(2!2!)= 6 ways, so
probability = 6* (1/66) = 1/11

The difficult way:
We subtract from 1 the cases of picking 1 #5 or none.
P(1-#5)= 4P1 * (1/12*10/11*9/10*8/9)
= 4* (4/33)
= 16/33
P(0-#5)=10/12*9/11*8/10*7/9
= 14/33
P(2-#5) = 1 - (P(1-#5)+P(0-#5)
=1- (30/33)
=3/33
=1/11

Why do we do it the difficult way?
In counting or probability, it is always a good idea to think of different ways of doing the same calculation, to explore different options, and to check our calculations.

To find the probability of picking 2 marbles numbered 5 out of 4 marbles drawn without replacement, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's calculate the total number of possible outcomes. Since we are picking 4 marbles without replacement, the number of possible outcomes is determined by the combinations of 12 marbles taken 4 at a time, which can be calculated using the formula:

C(12, 4) = 12! / (4! * (12 - 4)!) = 495

Next, we need to determine the number of favorable outcomes. In this case, we want to pick 2 marbles numbered 5 out of the 4 marbles drawn. We can select 2 marbles numbered 5 in C(5, 2) ways, which can be calculated as:

C(5, 2) = 5! / (2! * (5 - 2)!) = 10

Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes
= 10 / 495
≈ 0.0202

Therefore, the probability that 2 out of the 4 marbles drawn are numbered 5 is approximately 0.0202, or 2.02%.