Geoboard problem

[stevecowall]

Problem:
We are given a square geoboard with 4 pegs on a side. If we select 3 pegs at random, what is the probability that the 3 pegs are the vertices of a triangle?

My way of solving:
Because it is a square geoboard with 4 pegs so there are 16 (4*4) pegs. Because it is a random 3 pegs so it must be 3/16.

This is an advance problem so i think this solution is wrong ...

 

[tkhunny]

Any idea how you might proceed?

1) It cannot be a triangle if all three are colinear.

2) There are only three patterns to consider.

--- 2a) First point hits one of four corner pegs.
--- 2b) First point hits one of eight edge pegs.
--- 2c) First point hits one of four interior pegs.

Are we getting anywhere?

 

[lookagain]

Problem:
We are given a square geoboard with 4 pegs on a side.
If we select 3 pegs at random, what is the probability
that the 3 pegs are the vertices of a triangle?

My way of solving:
Because it is a square geoboard with 4 pegs so there are 16 (4*4) pegs.
Because it is a random 3 pegs so it must be 3/16.

\(\displaystyle \text{*** Spoilers *** ahead. First notice. Please do not read }\)
\(\displaystyle \text{ahead if you do not want to read a worked out method.}\)
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\(\displaystyle \text{*** Spoilers *** ahead. Last notice.}\)
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stevecowall,

there are 16 C 3 = 560 ways of picking exactly 3 distinct points (pegs) from
16 distinct points (pegs). In particular, this case involves 16 points in a
square lattice.

You want to know the total ways of choosing exactly 3 distinct points in
this square lattice, such that they lie on the same line. Subtract those
number of ways from the 16 C 3 ways. Then divide that difference
by 16 C 3 to get the probability that the 3 chosen distinct points
(pegs) form a triangle.

There are four rows, four columns, and the two longest diagonals with
4 points each in them, respectively. There are 4 C 3 ways in choosing
the points in each of these 10 lines of points. That makes 4 ways
multiplied by 10 = 40 ways.

And there are four shorter diagonals each with 3 points. Each of those
diagonals give 3 C 3 number of ways = 1 way to have the points lie on
the same line for each respective diagonal. That makes 1 way multiplied
by 4 = 4 ways.


40 ways + 4 ways = 44 ways (where the 3 chosen points do not form a triangle).


\(\displaystyle Then, \ the \ probability \ is \ \ \dfrac{560 - 44}{560} = \dfrac{516}{560}\)


\(\displaystyle This \ reduces \ to \ \ \dfrac{129}{140} \approx 0.92\)


\(\displaystyle So, \ there \ is \ about \ a \ 92\% \ probability \ that \ exactly \ 3 \ \)

\(\displaystyle distinct \ points \ chosen \ from \ 16 \ distinct \ points \ in \ a \)

\(\displaystyle 4 \ by \ 4 \ square \ lattice \ will \ be \ the \ vertices \ of \ a \ triangle.\)



I ask users to look at the problem in their own ways and see if they arrive
at the same answer as mine.

 

[soroban]

Hello, stevecowall!

I used Brute Force (listing and counting).

We are given a square geoboard with 4 pegs on a side.
If we select 3 pegs at random, what is the probability that the 3 pegs are the vertices of a triangle?

Choosing 3 pegs from the available 16, there are:.\(\displaystyle {16\choose3} \:=\:560\) possible choices.

How many of these are three collinear points?


Horizontal

There are 4 cases for a row-of-three:

. . \(\displaystyle \begin{array}{cccccccccc}\bullet\;\bullet\;\bullet\;\,\circ && \bullet\;\bullet\;\circ\;\,\bullet && \bullet\;\circ\;\bullet\;\,\bullet && \circ\;\bullet\;\bullet\;\,\bullet \end{array}\)

And we have a choice of 4 rows: .\(\displaystyle 4\cdot4 \,=\,16\) ways.


Vertical

There are 4 cases for a column-of-three:

. . \(\displaystyle \begin{array}{cccccccccc}\bullet &\quad& \bullet &\quad& \bullet &\quad& \circ \\ \bullet && \bullet && \circ && \bullet \\ \bullet && \circ && \bullet && \bullet \\ \circ && \bullet && \bullet && \bullet \end{array}\)

And we have a choice of 4 columns: .\(\displaystyle 4\cdot4 \,=\,16\) ways.


Diagonal

There are 4 "short" diagonals \(\displaystyle (\searrow)\):

. . \(\displaystyle \begin{array}{cccc}\bullet & . & . & . \\ . & \bullet & . & . \\ . & . & \bullet & . \\ . & . & . & . \end{array} \quad \begin{array}{cccc}. & \bullet & . & . \\ . & . & \bullet & . \\ . & . & . & \bullet \\ . & . & . & . \end{array} \quad \begin{array}{cccc}.&.&.&. \\ \bullet &.&.&. \\ . & \bullet &.&. \\ . & . & \bullet & . \end{array} \quad \begin{array}{cccc}.&.&.&. \\ .& \bullet &.&. \\ .&.& \bullet &. \\ .&.&.&\bullet \end{array}\)

And 4 more \(\displaystyle (\nearrow)\).


There are 2 "long" diagonals \(\displaystyle (\searrow).\)

. . \(\displaystyle \begin{array}{cccc}\bullet &.&.&. \\ .&\bullet &.&. \\ .&. & . &. \\ .&.&.&\bullet \end{array} \qquad \begin{array}{cccc}\bullet &.&.&. \\ .&.&.&. \\ .&.&\bullet&. \\ .&.&.&\bullet \end{array}\)

And 2 more \(\displaystyle (\nearrow).\)

Hence, there are:.\(\displaystyle 4 + 4 + 2 + 2 \,=\,12\) diagonals.


We have: .\(\displaystyle 16 + 16 + 12 \:=\:44\) sets of collinear points.

Hence: .\(\displaystyle 560 - 44 \,=\,516\) sets that can form a triangle.

Therefore: .\(\displaystyle P(\text{triangle}) \:=\:\dfrac{516}{560} \:=\:\dfrac{129}{140}\)