Here is the question I am struggling with:
A box has 16 Balls, of which 8 are Green, 6 are Red, and 2 are Blue. If you draw 2 Balls with replacement, what is the probability of getting 1 Green Ball and 1 Blue Ball in no particular order?
I see three different ways to get an answer to this problem. Please refute my wrong answers with explanations because I am confused.
Probability of getting one green: 8/16
Probability of getting one blue: 2/16
(8/16) * (2/16) = 1/16 final answer
Method 2: Since the question said order does not matter, I still figured order does count into this equation so I approached it by finding the probability that the green ball is selected first, then the blue ball. Then add that probability to selecting the blue ball first, then the green ball.
Probability of getting green first then blue: (8/16) * (2/16) = 1/16
Probability of getting blue first then green: (2/16) * (8/16) = 1/16
therefore, the final answer is 1/16 + 1/16 = 1/8.
Note: This confuses me because we are double counting the answer, the problem said that order does not matter, but why doesn't the Method 1 take this into account?
Method 3 (Combination Method):
There are Comb(16,2) possible ways to select 2 balls out of 16
There are Comb(8,1)*Comb(2,1) ways to select a green and a blue ball
Probability of one green and one blue = Comb(8,1)*Comb(2,1)/Comb(16,2) = 16/120 = 2/15 final answer
Which one of these, if any, is the correct answer? The book says it is 1/8, but can someone please explain more and explain why my other methods are wrong. Thanks!