[ASK] Probability with Factors

[Monoxdifly]

In a bag there are m white balls and n red balls with mn = 200 and there are more white balls than red balls. If two balls are taken randomly at once and the probability of taking two different colored balls is \(\displaystyle \frac{40}{87}\) then the value of 2m + 3n is ...
A. 30
B. 45
C. 50
D. 70
E. 80

Okay, so the possibility of m and n are like this:
m = 200 and n =1
m = 100 and n = 2
m = 50 and n = 4
m = 40 and n = 5
m = 25 and n = 8
m = 20 and n = 10

Do I need to count their probability one by one then adding them up to make \(\displaystyle \frac{40}{87}\)? Or am I not supposed to do that?

 

[skeeter]

P(two different colors) =

$\dfrac{m}{m+n} \cdot \dfrac{n}{m+n-1} + \dfrac{n}{m+n} \cdot \dfrac{m}{m+n-1} = \dfrac{40}{87}$

$\dfrac{2mn}{(m+n)(m+n-1)} = \dfrac{40}{87}$

$\dfrac{400}{k(k-1)} = \dfrac{40}{87}$, where $k = m+n$

$k = m+n = 30 \text{ and } mn = 200 \implies m = 20 \text{ and } n = 10$

 

[Monoxdifly]

Took me a while to understand that k = 30 comes from the factorization of \(\displaystyle k^2-k-870=0\), but thank you. Now I understand. :)