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 Segment Section Settlement (Posted on 2014-08-28)
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.

 No Solution Yet Submitted by K Sengupta Rating: 5.0000 (1 votes)

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 solution and verification | Comment 2 of 3 |
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
End If
ct = ct + 1
End If
Next
Next
Next
Next
Next
Next
Text1.Text = Text1.Text & Str(goodct) & Str(ct) & Str(goodct / ct)

460 / 3540 ~= .129943502824859

 Posted by Charlie on 2014-08-28 12:44:27

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