Let $S$ be the set of points whose coordinates $x$ and $y$ are integers that satisfy $0\le x\le3,$ $0\le y\le4$. Two distinct points .....

Question

Let $S$ be the set of points whose coordinates $x,$ $y$ are integers that satisfy $0\le x\le3,$ $0\le y\le4$. Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is?

My approach:

I tried finding the probability of the points not belonging to $S$ as that makes the problem easier.

Let the two points be $(a,b)$ and $(c,d)$

Since $a,b,c,d$ belong to $\mathbb{Z}$, their midpoint won't belong to $S$ if $\{a,c\}$ and $\{b,d\}$ are both: one odd one even. (Like $a$ is odd, $c$ is even or vice versa, same for $(b,d)$.)

Now, for selecting $a,c$:

$_2C_1$ (for selecting which one is odd and which is even) $\ast\, _2C_1 \ast\, _2C_1$ ( since for $x$ there are $2$ even numbers: $0,2$ and $2$ odd numbers: $1,3$)

Similarly for selecting $b,d$:

$_2C_1 \ast\, _3C_1 \ast\, _2C_1$ (since for $y$ there are $3$ even numbers and $2$ odd numbers)

Total favourable cases = $_2C_1 \ast\, _2C_1 \ast\, _2C_1 \ast\, _2C_1 \ast\, _3C_1 \ast\, _2C_1 = 96$(Multiplying both)

Total cases = $_{20}C_2 = 190$

Therefore probability = $96/190 = 48/95$

Required probability = $1 - 48/95 = 47/95$

But the given answer is $21/95$. Where did I go wrong?

Answer 1

Just count, for each possible midpoint, the number of pairs with that midpoint, which is not too hard to do manually:

0 1 2 1 0
1 4 7 4 1
1 4 7 4 1
0 1 2 1 0

So there are $42$ admissible pairs out of $\binom{20}2$ for a probability of $\frac{21}{95}$.