### Autor Tema: probability

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08 Noviembre, 2019, 07:36 am
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#### jacks

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Q: The sum of $3$ positive integer is $20.$ Then the probability that they form a Triangle is

i am assuming $a+b+c=20,a,b,c \in \mathbb{N}$ using wlog $a\leq b \leq c$

for triangle $a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10$ and $a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7$

so range of $7\leq c\leq 10$

For $c=7.$ we have $a+b=13,$ Then ordered pairs $(6,7)$

For $c=8.$ we have $a+b=12,$ Then ordered pairs $(6,6),(5,7),(4,8)$

For $c=9.$ we have $a+b=11,$ Then ordered pairs $(5,6),(4,7),(3,8),(2,9)$

For $c=10.$ we have $a+b=10,$ Then ordered pairs $(5,5),(4,6),(3,7),(2,8),(1,9)$

ordered pairs to form a triangle is $13$(favorable ways)

How do i calculate Total ways. please see it Thanks

08 Noviembre, 2019, 11:27 am
Respuesta #1

#### Luis Fuentes

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Hola

Q: The sum of $3$ positive integer is $20.$ Then the probability that they form a Triangle is

i am assuming $a+b+c=20,a,b,c \in \mathbb{N}$ using wlog $a\leq b \leq c$

for triangle $a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10$ and $a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7$

so range of $7\leq c\leq 10$

For $c=7.$ we have $a+b=13,$ Then ordered pairs $(6,7)$

For $c=8.$ we have $a+b=12,$ Then ordered pairs $(6,6),(5,7),(4,8)$

For $c=9.$ we have $a+b=11,$ Then ordered pairs $(5,6),(4,7),(3,8),(2,9)$

For $c=10.$ we have $a+b=10,$ Then ordered pairs $(5,5),(4,6),(3,7),(2,8),(1,9)$

ordered pairs to form a triangle is $13$(favorable ways)

How do i calculate Total ways. please see it Thanks

You can compute the number of all (unordered) triplets which form a triangle.

You have $5$ ordered triples of type $(a,a,b)$. There are $3$ ways of rearrange it.

And you have $8$ ordered triples of type $(a,b,c)$. There are $6$ ways of rearrange it.

So there are $5\cdot 3+8\cdot 6=63$ triplets (a,b,c) providing a triangle.

The total ways are the number of integer solutions of:

$x+y+z=20$ with $x,y,z\geq 1$

This is equivalent to the number of non negative integer solutions of:

$a+b+c=17$

that is $\displaystyle\binom{3+\color{red}17\color{black}-1}{3-1}$

Best regards.

CORRECTED

08 Noviembre, 2019, 12:08 pm
Respuesta #2

#### Richard R Richard • Ingeniero Industrial
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 a b c $\triangle$vs$\cancel{\triangle}$ order 1 1 18 $\cancel{\triangle}$ 3 2 2 16 $\cancel{\triangle}$ 3 3 3 14 $\cancel{\triangle}$ 3 4 4 12 $\cancel{\triangle}$ 3 5 5 10 $\triangle$ 3 6 6 8 $\triangle$ 3 7 7 6 $\triangle$ 3 8 8 4 $\triangle$ 3 9 9 2 $\triangle$ 3 1 2 17 $\cancel{\triangle}$ 6 1 3 16 $\cancel{\triangle}$ 6 1 4 15 $\cancel{\triangle}$ 6 1 5 14 $\cancel{\triangle}$ 6 1 6 13 $\cancel{\triangle}$ 6 1 7 12 $\cancel{\triangle}$ 6 1 8 11 $\cancel{\triangle}$ 6 1 9 10 $\triangle$ 6 2 3 15 $\cancel{\triangle}$ 6 2 4 14 $\cancel{\triangle}$ 6 2 5 13 $\cancel{\triangle}$ 6 2 6 12 $\cancel{\triangle}$ 6 2 7 11 $\cancel{\triangle}$ 6 2 8 10 $\triangle$ 6 3 4 13 $\cancel{\triangle}$ 6 3 5 12 $\cancel{\triangle}$ 6 3 6 11 $\cancel{\triangle}$ 6 3 7 10 $\triangle$ 6 3 8 9 $\triangle$ 6 4 5 11 $\cancel{\triangle}$ 6 4 6 10 $\triangle$ 6 4 7 9 $\triangle$ 6 5 6 9 $\triangle$ 6 5 7 8 $\triangle$ 6

There is not more ...I think so..
without ordering

$P_{\triangle} =\dfrac{13}{33}$

with order and rotation

$P_{\triangle} =\dfrac{3\cdot 5+6\cdot 8}{3\cdot 9+6\cdot 24}=\dfrac{63}{171}=\dfrac{7}{19}$
Saludos  $\mathbb {R}^3$

11 Noviembre, 2019, 03:22 pm
Respuesta #3

#### jacks

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12 Noviembre, 2019, 07:55 am
Respuesta #4

#### Luis Fuentes

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Hi

Note that I had a mistake. It has been corrected.

The total ways are the number of integer solutions of:

$x+y+z=20$ with $x,y,z\geq 1$

This is equivalent to the number of non negative integer solutions of:

$a+b+c=17$

that is $\displaystyle\binom{3+\color{red}17\color{black}-1}{3-1}$

Best regards.

CORRECTED

Best regards.