S is the set of points each of whose whose coordinates x, y and z is an integer that satisfy:
0 ≤ x ≤ 2, 0 ≤ y ≤ 3 and 0 ≤ z ≤ 4
Two distinct points are randomly chosen from S.
Determine the probability that the midpoint of the segment that they determine also belongs to S.
In order to get integral coordinates for the midpoint, the x coordinates of the end points must be either both even or both odd, and the same holds for the y and z coordinates.
There are 2 x values that are even and 1 that is odd.
There are 2 y values that are even and 2 that are odd.
There are 3 z values that are even and 2 that are odd.
The probability that both x values chosen are even is 4/9 and that both are odd is 1/9 for a total of 5/9.
The probability that both y values chosen are even is 1/4 and that both are odd is 1/4 for a total of 1/2.
The probability that both z values chosen are even is 9/25 and that both are odd is 4/25 for a total of 13/25.
These are independent so the overall probability of the necessary condition is the product 5/9 * 1/2 * 13/25 = 13/90.
However, the probability that the same point was chosen for both ends and has to be thrown out is 1/3 * 1/4 * 1/5 = 1/60.
But we can't just subtract 13/90 - 1/60 = 23/180 to get the overall probability, as the case where both points are coincident is excluded from the beginning.
P(hit given separate points) = P(hit and separate points)/P(Separate points)
= (23/180) / (59/60) = 23/177 ~= .1299435028248588
This is confirmed by a program taking all possibilities into account:
For x = 0 To 2
For y = 0 To 3
For z = 0 To 4
For xx = 0 To 2
For yy = 0 To 3
For zz = 0 To 4
If x <> xx Or y <> yy Or z <> zz Then
x3 = (x + xx) / 2
y3 = (y + yy) / 2
z3 = (z + zz) / 2
If x3 = Int(x3) And y3 = Int(y3) And z3 = Int(z3) Then
goodct = goodct + 1
ct = ct + 1
Text1.Text = Text1.Text & Str(goodct) & Str(ct) & Str(goodct / ct)
460 / 3540 ~= .129943502824859
Posted by Charlie
on 2014-08-28 12:44:27