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?
in this problem i am not getting how to proceed like what is sample space?it is quiet confusing to me. Please give any simple view for this problem.
Probability – Probability That Broken Sticks Will Not Form a Triangle
probabilityprobability theorypuzzle
Related Solutions
The three triangle inequalities are
\begin{align} x + y &> 1-x-y \\ x + (1-x-y) &> y \\ y + (1-x-y) &> x \\ \end{align}
Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.
(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is $$\int_0^{1/2}\frac1{2-2x}-\frac{2x-1}{2x-2}\,dx=\ln 2-\frac12\approx0.19314$$ I believe that's the answer you calculated.
Let $\lambda$ be the length of the bigger piece and we split it into two smaller pieces $\lambda\mu$ and $\lambda(1-\mu)$. It is clear $\lambda$ and $\mu$ are uniform random variables $\sim \mathcal{U}(\frac12,1)$ and $\mathcal{U}(0,1)$ respectively.
In order for the three pieces with lengths $\;1-\lambda, \lambda\mu, \lambda(1-\mu)\;$ to form a triangle, the necessary and sufficient conditions are the fulfillment of following three triangular inequalities:
$$\begin{cases}
\lambda \mu + \lambda (1-\mu) &\ge 1-\lambda\\
\lambda \mu + (1 - \lambda) &\ge \lambda (1-\mu)\\
\lambda (1-\mu) + (1-\lambda) &\ge \lambda \mu
\end{cases}
\quad\iff\quad
\begin{cases}
\lambda \ge \frac12\\
\mu \ge 1 - \frac{1}{2\lambda}\\
\frac{1}{2\lambda} \ge \mu
\end{cases}$$
The first inequality is trivially satisfied because we are told to break the
bigger piece.
The probability we seek is given by:
$$2\int_{1/2}^1 \int_{1-\frac{1}{2\lambda}}^{\frac{1}{2\lambda}} d\mu d\lambda = 2\int_{1/2}^1 \left(\frac{1}{\lambda} - 1 \right) d\lambda = 2\log 2 - 1 \approx 38.6294\%$$
Best Answer
Choose a point $x$ between $(0,0.5)$. Then this point will be the breaking point on the stick. $x$ will be the length of smaller segment. Then you break the bigger part $1-x$ which is the one you have to break by 4:3. This means that you get two other sticks with lengths $(1-x)\frac{4}{7}$ and $(1-x)\frac{3}{7}$. The conditions for having triangle will be: $$ (1-x)\frac{4}{7} < (1-x)\frac{3}{7}+x $$ $$ (1-x)\frac{3}{7} < (1-x)\frac{4}{7}+x $$ $$ x < (1-x)\frac{3}{7}+(1-x)\frac{4}{7} $$ All of them except the first one is automatically satisfied. For the first one you should have: $$ x>\frac{1}{8} $$ Now the probability will be equal to picking a point in $(0,0.5)$ bigger that $\frac{1}{8}$. Assuming uniform distribution you get: $$ \mathbb{P}(x>\frac{1}{8})=\frac{3}{4} $$ which gives you the probability of making a triangle so the inverse will be $\frac{1}{4}$.