2016 AMC 12A Problem 13

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}

N

grows large. What is the sum of the digits of the least value of

N

such that

P(N)\lt\dfrac{321}{400}?

Show solution

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),

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.