Probability of Ordering Balls in a Line (2016 AMC 10A #17)

2016 AMC 10A #17

Problem: 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 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 4/5 as N grows large. What is the sum of the digits of the least value of N such that P(N) < 321/400?

Generalized Problem: Let N be a multiple of odd positive integer M. One red ball and N green balls are arranged in a line in random order. Let P(N) be the probability that at least $\frac{\frac{M+1}{2}}{M}$ of the green balls are on the same side of the red ball. What is P(N) in terms of M?

Solution: Let L be $\frac{N}{M}$, or $N=L*M$. First, we must note that we can first place all the N green balls in a line, with N+1 places to place the red ball, with equal probability. Then, all we need to find is the number of invalid positions and subtract it from the total N+1 positions of the red ball to find the probability P(N).

In the diagram below with M = 5 and L = 6, we can see that there are M-1 invalid positions of the $N+1=L*M + 1$ total positions. So, $P(N) = \frac{(L*M+1)-(M-1)}{L*M+1} = \frac{L*(M-1) + 2}{L*M+1}$, which is our general formula for the problem.

Applying this to our original problem with M=5, we can see that we want the least L such that $\frac{4L + 2}{5L+1} < \frac{321}{400}$. We can simplify this to $400(4L + 2) <  321(5L+1)$ or $5L > 479$. Since the question asks for the least $N = 5L$, we have that $N = 480$, whose digits have a sum of $4+8+0=12$.

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