IMO C.

The probability of that both the drawn balls are green:

n/(5+n) * (n-1)/(5+(n-1)) >= 1/2
n/(5+n) * (n-1)/(n+4) >= 1/2
2n(n-1)>=(5+n)(n+4)
n^2-11n-20>=0

Substitute 13 as the value of n & you will see that 6>0. This is the minimum value (check 11 & 9 to see that the inequality does not satisfy).

Vemuri wrote:IMO C.

The probability of that both the drawn balls are green:

n/(5+n) * (n-1)/(5+(n-1)) >= 1/2
n/(5+n) * (n-1)/(n+4) >= 1/2
2n(n-1)>=(5+n)(n+4)
n^2-11n-20>=0

Substitute 13 as the value of n & you will see that 6>0. This is the minimum value (check 11 & 9 to see that the inequality does not satisfy).

The best way to solve this is to use the options. Even after framing the quadratic you were doing the same.

Also start with the median value so that based on the outcome of the result you can decide whether the answer is more or less than the option chosen by you.

Considering n=13 we get the required probability as

13/18 x 12/17 = 26/51 > 1/2

Now you have to check n =11 also before concluding the answer

If n =11, the required probab is

11/16 x 10/15 = 11/24 <1/2

Hence the least value of N is 13