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.

hot18594 Jun 30, 2022

#5**+1 **

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

#1**+1 **

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!

Guest Jun 30, 2022

#2

#3**+1 **

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

#4**0 **

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**+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