TIFR CSE 2013 | Part A | Question: 17

Question

A stick of unit length is broken into two at a point chosen at random. Then, the larger part of the stick is further divided into two parts in the ratio $4:3$. What is the probability that the three sticks that are left CANNOT form a triangle?

• A. $1/4$
• B. $1/3$
• C. $5/6$
• D. $1/2$
• E. $\log_{e}(2)/2$

Comments

https://en.wikipedia.org/wiki/Triangle_inequality

Answer 1

Let the stick be broken to $x$ and $l-x$ with $x$ being the larger part.

To not get a triangle, sum of two sides must be smaller than the third one.

$\implies (l-x) +3x/7 < 4x/7 \implies l < 8x/7 \implies x > 7l/8$.

It is given that $x >l/2$ ($x$ being the larger part) $\implies \left(7l/8,l\right)$ is the favorable case from $\left(l/2, l \right)$ which gives probability $= (1/8)/(1/2) = 1/4$.

Comments

For those who are thinking how $\frac{1}{2}$ became the sample space. It is because whenever largest length > ½, then the triangle is not possible.

So as per problem statement our smaller side must range from 0<l-x<½ . But doing the above calculations, we noticed that as per given conditions, it can only range from 0<l-x<1/8.

And hence the probability.

Answer 2

let us assume we divide string at length x such that x is bigger part i.e x>1/2 and smaller part is 1-x ,further we divide x into 2parts of ratio 4:3 means 1 part is 4x/7 and second part is 3x/7 so far we have establish the value of each side (a,b,c) in term of x (4x/7,3x/7,1-x)

now for triangle to form following 2 property must satisfies a+b>c  and |a-b|<c for any side (a,b,c) now we have to permute a b c as per above to find value of x

1)a=4x/7 b=3x/7,c=1-x      ----- x belong to (1/2 ,7/8)

2)a=4x/7 b=1-x ,c=3x/7     ----- x < 7/6 futile because x cannot be greater than 1

3)a=3x/7 b=1-x ,c=4x/7     ----- x belong to (1/2 ,7/6)

4)a=3x/7 b=4x/7 ,c=1-x     ----- x belong to (1/2 ,7/8)

5)a=1-x b=4x/7 ,c=3x/7     ----- x < 7/8

6)a=1-x b=3x/7 ,c=4x/7     ----- x < 7/6 futile because x cannot be greater than 1

so finally x should be between 1/2 and 7/8 inorder to form triangle hence it can take (7/8-1/2) =3/8 values therefore

probabilty of forming a triangle is 3/8(favourable values)/ 1/2(total values)=3/4

probability of not forming a triangle will be 1-3/4 = 1/4    

Answer 3

Ans is 1/6

Here the stick divided into two halves

Next the larger part divided into 4:3=4x:3x

we know for triangle a+b>c

a=3

c=4

3+b>4

so b cannot be 1 for making a triangle

now, given larger edge  divided into 4x:3x

so larger edge length 4x+3x=7x

now, value of smaller edge can be upto 6x

among which it cannot be 1  i.e. 1x

so probablity of not making a triangle= 1/6

Comments

for triangle a^2+b^2>=c^2

This is for right angled triangle only. For a general triangle the rule is sum of any two sides > the other side. 

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yes corrected now.

sum of any two sides = the other side, then three points must be collinear

and for making triangle sum of any two sides must be greater than the other side. . 

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Now you have to change discrete probability to continuous probability- the smaller piece can come from two sides of the stick also.

Answer 4

Let $x$ be the smaller length and the larger length be $7$.

First we will calculate the probability of the favorable condition for making a triangle. As the larger length $7$ is divided into the ratio $4:3$. It means to make a triangle, the following property should hold

$|a-b|<x<a+b \\\Rightarrow |4-3|<x<4+3\\\Rightarrow 1<x<7$

Here since a length is a uniform continuous quantity, hence we use discrete terminals to calculate the probability like from set $\{1,2,3,4,5,6,7\}$

So the probability of forming a triangle using ($x$, $4$, $3$) is

$\begin{align}p&=\frac{\text{Total favourable lengths of }x}{\text{(Total possible lengths of }x)+1~(\text{only one larger length as }7)}\\&=\frac{7-1}{7+1};~[\scriptsize\text{It's } (b-a) \text{ as favourable lengths (actually continuous) for }a<x<b]\\&=\frac{6}{8}\\&=\frac{3}{4} \end{align}$.

$\therefore$ The probability for not making a triangle $=1-p=1-\frac{3}{4}=\frac{1}{4}$.

 

So the correct answer is A.

Answer 5

It is given that we have unit length stick, so let the stick be broken at point $(x,0)$ supposing that stick is placed horizontally from origin $(0,0)$. Such that one part would be having length of $x$ units and another would be having length of $1-x$ units. We are instructed to further break the largest part of stick in two pieces. 

Let, $x$ length be the largest part, such that it would be broken into $\frac{3x}{7}$ and $\frac{4x}{7}$ unit lengths. Now we have stick broken into three pieces of length given in $3$-ary tuple $(1-x, \frac{3x}{7}, \frac{4x}{7})$. 

If $a$, $b$, and $c$ are lengths of the sides of the triangle, with no side being greater than $c$, then the triangle inequality states that $a+b\ge c$. 

$\underline{\textbf{Observations}}$

To satisfy triangle inequality, no side of the triangle should be greater than $\frac{1}{2}$ unit length. (Why?)

So, length of any side of the triangle should lie in $(0,\frac{1}{2})$. 

Keeping the above observation in mind, we can say that length of any side of triangle should be in $[\frac{1}{2}, 1]$. (Since, we need to find for the case when triangle CANNOT be formed).

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We want the values of $x$ for which the triangle inequality is not satisfied. So, we will find such values in cases.

$\textbf{Case 1:} \ (1-x) + \frac{3x}{7} < \frac{4x}{7} \implies x>\frac{7}{8}$ (in $(\frac{7}{8}, 1])$

$\textbf{Case 2:} \ (1-x) + \frac{4x}{7} < \frac{3x}{7} \implies x>\frac{7}{6}$ (Not Possible, Out of Domain)

$\textbf{Case 3:} \ \frac{3x}{7} + \frac{4x}{7} < (1-x) \implies x<\frac{1}{2}$ (Not Possible, Out of Domain) 

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$$P(x>\frac{7}{8}) = 1-P(x \le \frac{7}{8}) = 1- \int_{\frac{1}{2}}^{\frac{7}{8}}\underbrace{\dfrac{1}{1-\frac{1}{2}}}_{pdf}dx = \dfrac{1}{4}$$

$\textbf{Option (A) is correct}$