A stick has a length of $5$ units. The stick is then broken at two points chosen at random. To find the probability that all three resulting pieces are shorter than $3$.
I tried with complement prob: We can have only one piece $\ge 3$. So I tried to find how many ways we can break the remaining into two parts.
On
Hint: try a graphic approach. You can plot all possible "breaks" as the square $[0,5]\times[0,5]$. Then, try to think about what the length of the three pieces is for each point in that square, and what condition must be met so that the longest piece is shorter than $3$.
Hint 2: It might be easier to look at the triangle $\{(x,y)\in[0,5]\times[0,5]: x \leq y\}$ and then its complement, and examine the condition on each of them separately. Or even just notice that it is enough to examine one of the triangles as the other one is a mirror image of the first.
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Lets say that the pieces are $x,y,z$ where $x+y+z =5$.We want to find the cases that all pieces are less than or equal to $3$ , (remember that because of this is continuous probability distribution , it does not matter whether it is $<$ or $\leq$).
Now , lets randomly work over the pieces , we want that $z \leq 3$ and $x+y+z =5$ , to satisfy this condition , when $z \leq 3$ , $x+y \geq2$. So ,we can conclude that if we find the probability that $x+y \geq 2$ , we can find the result , by the way realize that $x \leq5 ,y\leq5$ in our sample set.
Finding the probability is equal to finding the area between the lines $x+y =5$ and $x+y=2$. Then it is equal to $$\frac{5 \times 5}{2} - \frac{2 \times 2}{2} =\frac{21}{2}$$ .Moreover , our sample set is the points where $x+y =5$ , then it is equal to $\frac{5 \times 5}{2}$
Then , the probability is $$\frac{\frac{21}{2}}{\frac{5 \times 5}{2}}= \frac{21}{25}$$
$\mathbf{\text{EDITION:}}$ After we find $x+y \geq 2$ , we must also add the restrictions for $x$ and $y$ such that $x \leq3 ,y\leq3$ , these restrictions will construct two gaps between the lines of $x+y \geq 2$ and $x+y \leq 5$ , the area of each of these gaps is $\frac{2\times2}{2}$ i so we must subtract them from the total. Then , $$\frac{25}{2}-\frac{4}{2}-\frac{4}{2}-\frac{4}{2} = \frac{13}{2}$$
So , $$\frac{\frac{13}{2}}{\frac{25}{2}}= \frac{13}{25}$$
On
The probability the first piece is too long is the probability both $x$ and $y$ are greater than $3$, which is $\left(\frac{5-3}{5}\right)^2=\frac4{25}$
The probability the central piece is too long is the probability $|x-y|>3$, double the probability $x>y+3$, so also $2\times \frac12 \left(\frac{5-3}{5}\right)\left(\frac{2-0}{5}\right)= \frac4{25}$
The probability the last piece is too long is the probability both $x$ and $y$ are less than $2$, which is also $\left(\frac{2-0}{5}\right)^2=\frac4{25}$
You cannot have more than one of the pieces too long, since $3+3>5$. So the probability none of the pieces are too long is $1-\frac{12}{25}=\frac{13}{25}$
Suppose we cut the stick at $x$ and $y$:
In region A, $y>x+3$, so the middle piece is longer than $3$.
In region B, $x>3$ and $y>3$, so the leftmost piece is longer than $3$.
In region C, $x<2$ and $y<2$, so the rightmost piece is longer than $3$.
In region D, $x>y+3$, so again the middle piece is longer than $3$.
The areas of these four regions add up to $\dfrac{12}{25}$. So the probability that every piece is shorter than $3$ is $\dfrac{13}{25}$, as Henry says.