bluementor wrote:no. of red marbles = x
no. of blue marbles = y
prob(different colors) = prob(1st = red, 2nd = blue) + prob(1st = blue, 2nd = red)
= prob(1st = red)*prob(2nd = blue) + prob(1st = blue)*prob(2nd = red)
= (x/(x+y))*(y/(x+y-1)) + (y/(x+y))*(x/(x+y-1))
= xy/(x+y)(x+y-1) + xy/(x+y)(x+y-1)
= 2xy/(x+y)(x+y-1)
given that, prob(different colors) = 1/2, so,
1/2 = 2xy/(x+y)(x+y-1)
(x+y)(x+y-1) = 4xy
x^2 + y^2 + 2xy - x - y = 4xy
x^2 - 2xy + y^2 = x + y
(x - y)^2 = x + y
x can be 7, 9, 11, 13, and 15, and y must be an integer. The only way forward that I can think of is to plug in various values for x and see if there is an integer solution for y. Trying out choices A to E, you will see that only choice E yields an integer solution for y (x = 15, y = 5).
Choose E.
P.S. I'd love to see a more elegant solution to this problem.
-BM-
Here is the OA
Let r be the number of red marbles and b the number of blue marbles. The number of ways of choosing two marbles is (r+b)C2 . The number of ways of choosing two different color marbles is rb.
Hence, we have 1/2= rb/(r+b)C2 = 2rb/(r + b)(r + b − 1)
.
This can be rewritten as
b2 − (2r + 1)b + r2 − r = 0.
Since the number of blue marbles must be an integer that is a root of the above equation, we deduce
that the discriminant of this quadratic is a square. In other words, there is an integer m such that
(2r + 1)2 − 4(r^r − r) = m2 =) 8r + 1 = m2.
The only r = {7, 9, 11, 13, 15} for which this equation holds for some integer m is 15, so the answer is 15. One can check that if r = 15 and b = 10, the probability that two randomly chosen marbles have different olors is in fact 0.5.
My solution:
In my approach I reasoned that the only number among the choices that is a factor of 5 is 15 so I would begin with 15 and test. If this did not work I would have to guess the answer.
15C1 x BC1/(15 +B)C2 = 1/2
After simplifying this gives
B^2 -41B +210 =0
which produces a discriminant that has a perfect square (841)^.5 =29
Once I recognize this perfect sqaure I choose E
