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

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The discussion focuses on calculating the probability that the midpoint of two randomly selected points from set S, defined by integer coordinates within specified ranges, also lies within set S. The set S includes points with x-coordinates from 0 to 2, y-coordinates from 0 to 3, and z-coordinates from 0 to 4, resulting in a total of 30 points. The key to the solution involves determining the conditions under which the midpoint remains an integer point within the defined ranges. Members kaliprasad and Opalg provided correct solutions, highlighting the importance of understanding midpoint calculations in discrete sets. The problem emphasizes combinatorial reasoning and probability in a geometric context.
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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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Congratulations to the following members for their correct solution:):

1. kaliprasad
2. Opalg

Solution from Opalg:
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 $p = \dfrac{23}{177}.$
 
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