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# A 10-cm stick has a mark at each centimeter. By breaking the stick at two of these nine marks at random, the stick is split into three piece

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A 10-cm stick has a mark at each centimeter. By breaking the stick at two of these nine marks at random, the stick is split into three pieces, each of integer length. What is the probability that the three lengths could be the three side lengths of a triangle? Express your answer as a common fraction.

Jun 30, 2022

### Best Answer

#5
+118577
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Hi Chris, hot18594, Ginger and guest.    And thanks for your answers and interaction

I think that we are all in agreement that there are 36 different possible combinations of cuts.

9 places to cut, cut in 2 places   9C2 = 36

We are also in agreement that the sides of the triangle will be (2,4,4) or (3,3,4) units in length.

BUT where can the CUTS happen for this??

I will look at 2,4,4

1 cut 3,4,5, cut, 7,8,9,10     that would give 2,4,4

or

1,2,3,cut,5,cut,7,8,9,10      that would give 4,2,4

or

1,2,3,cut,5,6,7,cut,9,cut      that would give 4,4,2

So there are 3 ways to cut the stick to get a trinagle of sides  2,4,4

there will also be 3 ways to cut the stick to get  3,3 and 4

So there are 6 favourable ways to cut the stick.

Probability of being able to make a triangle will be    6/36   =    1/6

I think that is what Ginger was refering to.

Jul 2, 2022

### 5+0 Answers

#1
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There are two ways to choose the lengths to be trianlges, (2,4,4) and (3,3,4).  The number of ways of choosing the break points is C(9,2) = 36.  So the probability is 2/36 = 1/18.  Hope that helps!

Jun 30, 2022
#2
+12
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Got it, its 1/6

Jun 30, 2022
#3
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Could  you explain your reasoning, hot18594  ???

I know  that one  website  claims that there are 6 triangles (3 each)  formed from sides of ( 3,3,4)  and (4, 4, 2) but these triangles are isosceles so any orientation of them is the same

I agree with the guest  ⇒   1 /  18

CPhill  Jun 30, 2022
edited by CPhill  Jun 30, 2022
#4
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Hello CPhill,

When calculating a statistic on indistinguishable elements or arraignments, they are (usually) treated as if they are distinguishable.  Though not distinguishable as one from another, their appearance is proportional to their multiplicity, and this affects their appearance counts in statistical measures.

GA

--. .-

GingerAle  Jun 30, 2022
#5
+118577
+1
Best Answer

Hi Chris, hot18594, Ginger and guest.    And thanks for your answers and interaction

I think that we are all in agreement that there are 36 different possible combinations of cuts.

9 places to cut, cut in 2 places   9C2 = 36

We are also in agreement that the sides of the triangle will be (2,4,4) or (3,3,4) units in length.

BUT where can the CUTS happen for this??

I will look at 2,4,4

1 cut 3,4,5, cut, 7,8,9,10     that would give 2,4,4

or

1,2,3,cut,5,cut,7,8,9,10      that would give 4,2,4

or

1,2,3,cut,5,6,7,cut,9,cut      that would give 4,4,2

So there are 3 ways to cut the stick to get a trinagle of sides  2,4,4

there will also be 3 ways to cut the stick to get  3,3 and 4

So there are 6 favourable ways to cut the stick.

Probability of being able to make a triangle will be    6/36   =    1/6

I think that is what Ginger was refering to.

Melody Jul 2, 2022