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In my closet, I have only hats that are green and hats that are blue. If I were to randomly choose two hats from the closet, it is equally probable
that they would be the same colour as it is that they would be different colours.
A friend asks me “What is the probability that if you were to randomly choose two hats from your closet that both hats would be green?”
My response is, “It is equal to the probability that if I instead randomly choose one hat from the closet,
that hat will be blue.” How many blue hats do I have? (Assume the number of hats of each colour is greater than zero)
Let G be the number of green hats and B be the number of blue hat.
The probability that two randomly chosen hats are both green is P(GG) = .
The probability that two randomly chosen hats are both blue is P(BB) = .
The probability that two randomly chosen hats are the same color is
P(GG) + P(BB) = + .
The probability that two randomly chosen hats are of different color is
P(GB) + P(BG) = + = .
So, your first equation, from the condition, is
P(GG) + P(BB) = P(GB) + P(BG), or + = .
Canceling common denominators, you get this equation in equivalent form
G*(G-1) + B*(B-1) = 2GB. (1)
We completed with the first part of the condition, and now start working with the second part.
From the second part, you have this equation
P(GG) = P(B), or = .
After canceling common factors in the denominators, it takes the form
= B, or, equivalently,
G*(G-1) = B*(G + B - 1) (2)
So, now our task is to solve the system of equations (1) and (2).
For it, replace G*(G-1) in the left side of (1) by B*(G+B-1), based on (2). You will get then
B*(G + B - 1) + B*(B-1) = 2GB.
Cancel B in both side
G + B - 1 + B - 1 = 2G, or
2B - 2 = G. (3)
Based on (3), replace G in (2) by 2B-2. You will get
(2B-2)*(2B - 2 -1) = B*((2B-2) + B -1).
Simplify it step by step
(2B-2)*(2B-3) = B*(3B-3)
4B^2 - 4B - 6B + 6 = 3B^2 - 3B
B^2 - 7B + 6 = 0
(B-6)*(B+1) = 0
Only positive root B = 6 makes sense.
ANSWER. 6 blue hats and 2B-2 = 2*6-2 = 10 green hats.
Solved.