Probability of Midpoint in Set S | POTW #307 Mar 28th, 2018

[anemone]

Here is this week's POTW:

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Let $S$ be the set of the points whose the $x$, $y$ and $z$ coordinates are integers that satisfy $0\le x \le 2$, $0\le y \le 3$ and $0\le z \le 4$. Two different points are randomly picked from $S$. What is the probability that the midpoint of the two picked points also belongs to $S$?

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[anemone]

Congratulations to the following members for their correct solution:):

1. kaliprasad
2. Opalg

Solution from Opalg:

Spoiler

Let $A$ and $B$ be randomly chosen points in $S$ and let $M$ be their midpoint. The condition for $M$ to be in $S$ is that its coordinates should all be integers. Therefore each of the three coefficients of $A$ and $B$ should be either both even or both odd.

For the $x$-coordinate, the possible values are $0$, $1$ and $2$. So the probability of the $x$-coordinate being even is $\frac23$, and the probability of it being odd is $\frac13$. Thus the probability that the $x$-coordinate of $M$ is an integer is $\bigl(\frac23\bigr)^2 + \bigl(\frac13\bigr)^2 = \frac59.$

For the $y$-coordinate, the possible values are $0$, $1$, $2$ and $3$. So the probabilities of it being even/odd are $\frac12$ both times, and the probability that the $y$-coordinate of $M$ is an integer is $\bigl(\frac12\bigr)^2 + \bigl(\frac12\bigr)^2 = \frac12.$

For the $z$-coordinate, the possible values are $0$, $1$, $2$, $3$ and $4$. So the probabilities of it being even/odd are $\frac35$ and $\frac25$, and the probability that the $z$-coordinate of $M$ is an integer is $\bigl(\frac35\bigr)^2 + \bigl(\frac25\bigr)^2 = \frac{13}{25}.$

Those three probabilities are independent, so the probability that $M$ is in $S$ is $\frac59 \cdot\frac12 \cdot\frac{13}{25} = \frac{13}{90}$.

But that is not the answer to the problem, because it ignores the given information that the points $A$ and $B$ are different. There are $3\cdot4\cdot5 = 60$ points in $S$, so there is a probability $\frac1{60}$ that $A=B$. In that case, $M=A=B$, so $M$ will certainly be in $S$. Let $p$ be the probability that $M$ is in $S$ given that $A$ and $B$ are distinct. Then it follows that $$ \tfrac{13}{90} = \tfrac1{60} + \tfrac{59}{60}p,$$ from which $p = \dfrac{23}{177}.$