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?

2 answers

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.