Question

Let $N$ be a positive multiple of $5.$ One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\dfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\dfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N)\lt\dfrac{321}{400}?$

$12$

$14$

$16$

$18$

$20$

Solution:

Write $N=5k.$ Number the positions of the red ball $0,1,\ldots,5k$ from one end; there are $5k+1$ equally likely positions.

Fewer than $\dfrac{3}{5}$ of the green balls lie on each side exactly when the red ball is in one of the positions $2k+1,2k+2,\ldots,3k-1,$ which is $k-1$ positions. Hence $P(N)=1-\dfrac{k-1}{5k+1}=\dfrac{4k+2}{5k+1}.$

Solving $\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}$ gives $400(4k+2)\lt 321(5k+1),$ so $1600k+800\lt 1605k+321$ and $5k\gt 479,$ meaning $k\gt 95.8.$ Thus $k=96$ and $N=480,$ whose digit sum is $4+8+0=12.$

Thus, the correct answer is A.

2016 AMC 12A Problem 13

Solution:

Problem 13 in Other Years