A box has 5 red balls and N blue balls. Two of the balls are randomly chosen. The probability that they are the same color is 11/21. Find the sum of all possible values of N

Answer 1

7CN = 7!/N!(7-N)! = 21 = 7!/2!5! = 7(6)/2 = 21 for denominator with N=2

ways to get 2 of same color = 11 = numerator

or add 10 blue to 5 red 15C10 = 15!/5!10!

= 15(14)(13)(12)11/5(4)(3)2 = 3(14)(13)(2) = 21(52)

and 52(11) ways to get 2 of same color

Answer 2

Probability both are red = (5/(N+5))(4/(4+N)) = 20/((N+5)(N+4))

Probability both are blue = (N/(N+5))((N-1)/(5+N-1)) = (N²-N)/((N+5)(N+4))

Probability both same color

20/((N+5)(N+4)) + (N²-N)/((N+5)(N+4)) = 11/21

doing the algebra

10N²-120N+200 = 0

N²-12N+20 = 0

(N-2)(N-10) = 0

N = 2,10

Sum: 2+10 = 12

Answer 3

We have N + 5 total balls, meaning we have _N+5C₂ = (N+5)(N+4) / 2 distinct pairs of balls that may be chosen.

Of those pairs, we have ₅C₂ = 10 distinct pairs of red (same color) balls and _NC₂ = N(N-1) / 2 distinct pairs of blue (same color) balls.

[10 + N(N-1)/2] / [(N+5)(N+4)/2] = [N² - N + 20] / [N² + 9N + 20] = 11/21

21N² - 21N + 420 = 11N² +99N + 220

10N² - 120N + 200 = 0

10(N - 2)(N - 10) = 0

N = 2 or N = 10 and the sum of these 2 possible answers is 12.