A bag contains `b` blue balls and `r` red balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. Then (a) `b+r=9` (b) `b r=18` (c) `|b-r|=4` (d) `b//r=2`

Options

A. b + r = 9
B. br = 18
C. `|b - r| = 4`
D. `b//r = 2`

Answer

A, B

Explanation

Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls is
`p_(1) = (.^(r )C_(2))/(.^(r+b)C_(2)) = (r(r-1))/((r+b)(r+b-1))`
The probaility of drawing two blue balls is
`p_(2) = (.^(b)C_(2))/(.^(r+b)C_(2)) = (b(b-1))/((r+b)(r+b-1))`
The probability of drawing one red and one blue ball is
`P_(3) = (.^(r)C_(1)xx.^(b)C_(1))/(.^(r+b)C_(2))=(2br)/((r+b)(r+b-1))`
By hypothesis, `p_(1)=5p_(2)` and `p_(3) = 6p_(2)`. Therefore,
r(r-1) = 5b(b - 1) and 2br = 6b(b-1)
implies r = 6, b = 3