Shendy's Uniform

Use geometric probability: Draw a square with side-length 10. We can bound our desired region by solving when $x+2y=8$ $x+2y=20$ If $x+2y=8$ , $y=-\dfrac{1}{2}x+4$ . We can draw this line on our square, and we cut off a triangular region with area $\dfrac{1}{2}\cdot4\cdot 8=16$ .

If $x+2y=20$ , $y=-\dfrac{1}{2}x+10$ . This line also cuts off a triangle, with area $\dfrac{1}{2}\cdot 5\cdot 10=25$ .

It is easy to see these regions do not intersect, and that the remaining area satisfies the inequality, so the area of the region is $100-16-25=59$ , and the answer is $59+100=\boxed{159}$ .

Define some new variables to make things simpler: $X = x, Y = 2y, Z = X + Y$ We also have the probability distribution $f_{\kappa}$ which has the property: $P_{\kappa}(a \leq k \leq b) = \int_{t=a}^{t=b} f_{\kappa}(t)dt$

The conditions of the problem give: $f_{X}(t) = \begin{cases} 1/10, & \mbox{if } 0 \leq t \leq 10 \\ 0, & \mbox{otherwise } \end{cases}$ $f_{Y}(t) = \begin{cases} 1/20, & \mbox{if } 0 \leq t \leq 20 \\ 0, & \mbox{otherwise } \end{cases}$

Now we have to work out $f_{Z}$ . If we think about, for example, $P_{Z}( 4 \leq z \leq 4 + \epsilon )$ , then we have to sum over all cases: $X = 0, Y = 4$ ; $X = \delta, Y = 4 - \delta$ ; etc. So we get: $f_{Z}(z) = \int_{t=-\infty}^{t=\infty} f_{X}(t) f_{Y}(z - t) dt$

This is actually known as the $\href{http://en.wikipedia.org/wiki/Convolution}{convolution}$ $f_{X} * f_{Y}$ , and the linked page has a proof that $\int_{-\infty}^{\infty} (f_{X} * f_{Y})(t)dt = (\int_{-\infty}^{\infty} f_{X}(t)dt)(\int_{-\infty}^{\infty} f_{Y}(t)dt)$ , making the convolution also a probability distribution.

To evaluate this we first note that $f_{X}(t)$ is only valid for $0 \leq t\leq 10$ , and since $f_{X}(t) = 1/10$ on this range we can evaluate it immediately, leaving: $f_{Z}(z) = \frac{1}{10} \int_{t=0}^{t=10} f_{Y}(z - t) dt$

We can simplify this with the change of variable $s = z - t$ , which means $ds = -dt$ giving: $f_{Z}(z) = \frac{1}{10} \int_{s=z-10}^{s=z} f_{Y}(s) ds$

Now we can evaluate this in a piecewise manner, bearing in mind that $f_{Y}(s) = 0$ outside the range $0 \leq s \leq 20$ ; and inside that range the antiderivative of $f_{Y}(s) = \frac{1}{20}$ is $\frac{s}{20}$ :

Case 1: $0 \leq z \leq 10$ : $f_{Z}(z) = \frac{1}{10} [\frac{s}{20}]_{0}^{z} = \frac{z}{200}$

Case 2: $10 \leq z \leq 20$ : $f_{Z}(z) = \frac{1}{10} [\frac{s}{20}]_{z-10}^{z} = \frac{10}{200}$

Case 3: $20 \leq z \leq 30$ : $f_{Z}(z) = \frac{1}{10} [\frac{s}{20}]_{z-10}^{20} = \frac{(30-z)}{200}$

Now we are in a position to calculate $\begin{aligned} P(8 \leq z \leq 20) &= \int_{8}^{20} f_{Z}(t) dt \\ &= \int_{8}^{10} \frac{t}{200} dt + \int_{10}^{20} \frac{10}{200} dt \\ &= [\frac{t^2}{400}]_{8}^{10} + \frac{1}{2} \\ &= \frac{9}{100} + \frac{50}{100} \\ &= \frac{59}{100} \end{aligned}$ so the required solution is $59 + 100 = \boxed{159}$

Footnote: I'd encourage the reader to draw the graph of $f_{Z}$ . The integrals could then be done more quickly based on the graph, by considering the area under the rectangular part, and the area under the trapezoidal part.

I suspect when I click "Post" that there will be some short solutions that make a lot of assumptions (e.g. giving $f_{Z}$ piecewise directly, assuming the steps leading up to it are 'obvious').

However this was my first time working on this sort of problem, so I hope that someone finds my full working useful, especially if they ever need to consider non-uniform probability distributions, or the convolution of more than two distributions (which is non-trivial, the case of just 2 is special).

Note : $P(A,B)$ denotes the probability of $A$ , given that $B$ is true.

By the inclusion-exclusion principle,

$P(8 \leq x+2y \leq 20) = 1 - P(x+2y < 8) - P(x+2y > 20).$

We know that

$P(x<8)=\frac{8}{10}=\frac{4}{5}$ and $P(y < \frac{8-x}{2} \mid E(x \mid x < 8)=4) = P(y<2)=\frac{2}{10}=\frac{1}{5},$ so $P(x+2y<8) = \frac{4}{5} \cdot \frac{1}{5} = \frac{4}{25}.$

Furthermore, we know that

$P(x+2y > 20) = P(y>\frac{20-x}{2} \mid E(x)=5) = P(y>\frac{15}{2}) = \frac{1}{4}.$

Therefore,

$\begin{aligned} P(8 \leq x+2y \leq 20) &= 1 - P(x+2y < 8) - P(x+2y > 20)\\ &= 1 - \frac{4}{25} - \frac{1}{4}\\ &= \frac{59}{100}, \end{aligned}$

so $a+b=59+100=\boxed{159}$ .

The region of possible $(x,y)$ is a square with vertices $(0,0), (10,0), (10,10), (0,10)$ . This region has area $10 \cdot 10 = 100$ .

The sub-region where $x + 2y < 8$ is the right triangle with vertices $(0,0), (8,0), (0,4)$ , which has area $\tfrac{1}{2} \cdot 8 \cdot 4 = 16$ .

Hence, $P[x+2y < 8] = \tfrac{16}{100}$ .

The sub-region where $x + 2y > 20$ is the right triangle with vertices $(0,10), (10,0), (10,10)$ , which has area $\tfrac{1}{2} \cdot 10 \cdot 5 = 25$ .

Hence, $P[x+2y > 20] = \tfrac{25}{100}$ .

Thus, $P[8 \le x+2y \le 20] = 1 - P[x+2y < 8] - P[x+2y > 20] = \tfrac{59}{100}$ .

Since $\gcd(59,100) = 1$ . the answer is $59+100 = \boxed{159}$ .

We draw a graph. The two lines described by $8 \le x + 2y \le 20$ are $y = 4 - \dfrac {x}{2}$ and $y = 10 - \dfrac {x}{2}$ . We want the area in between those two lines such that $0 \le x, y \le 10$ . To find this, we find the total area between the lines and subtract the area where $x$ exceeds $10$ .

The total area is the area inside the smaller triangle and outside the larger triangle. We subtract and we get: $A_{\text{between}} = \dfrac {10 \cdot 20}{2} - \dfrac {4 \cdot 8}{2} = 84.$ The bogus area is a triangle of horizontal leg $10$ and vertical leg $10 - \dfrac {10}{2} = 5$ . We subtract $25$ , so we get $59$ .

The total area at hand is $100$ , so the probability is $\dfrac {59}{100}$ , and our answer is $\boxed {159}$ .

Representing the sample space on the x-y plane as a 10x10 square the region we want is everything apart from a right angled triangles with vertices (0,0),(0,4) and (8,0) and (0,10),(10,5) and (10,0) with areas 16 and 25 respectively. The remaining area is 59 out of 100 so the probability we require is $\frac{59}{100}$

Graphing the lines 8 = x + 2y and x + 2y = 20, we the two lines intersect the the region bounded by x = 0, x = 10, y = 0, y = 10 at the points (8, 0), (0, 4), (0, 10), and (10, 5). Since the region we are looking for is everything in between 8 = x + 2y and x + 2y = 20, we find the area of this region using shoelace theorem on the points (8, 0), (0, 4), (0, 10), and (10, 5) as well as (10, 0). Doing so gives us an area of 59. The entire area of the region is 100, so 59/100 = a/b.

Thus, a + b = 59 + 100 = 159 .

I can't draw a picture, but I'll try my best to explain it:

To solve this, we use geometric probability. Let's graph the region between $0\le y \le10$ and $0 \le x \le10$ and see that the area of that figure is $100$ .

Then we graph $8 \le x+2y \le 20$ that fits inside the area we found above. The vertices of the resulting polygon are $\{(0,10), (0,4), (8,0), (10,0), (10,5)\}$ . We have to find the area, which is fairly simple, to get $59$ . Therefore probability is $\frac{59}{100}$ which means the answer is $\boxed{159}$

Draw the picture of the $10\times 10$ box and the corresponding lines. The desired area can be found by subtracting the are of the triangles outside of the region between lines from the are of the larger box. That is: $A = 10\times10-\frac12 (10\times 5) - \frac12 (4 \times 8) = 59$ The probability that a point chosen falls within that region is simply the ratio of that area to the overall area of the box. That is, the probability is $\dfrac{59}{100}$ . The answer must then be $100+59=159$

The problem can be viewed as ratio of area $\frac{between x + 2y = 8 , x+ 2y = 20}{square of length =10}$

first draw the graph. Note that favoured area = 50 + 25 -16 = 59 units, total possible area = 100 units, hence P = 59/100 ,giving a+b = 59 +100 =159

The question is basically to draw a nice picture. You have a square of 10 by 10 with given $x$ - and $y$ -coordinates. If any point $(x,y)$ satisfies the inequality, it lies within this square and between the lines $y=4-\frac{x}{2}$ and $y=10-\frac{x}{2}$ . The desired ratio, is the quotient of this area by the area of the square, which is exactly $\frac{59}{100}$ . Therefore the final answer is 159.

With 2d problems it often easiest to plot a graph. Here the possible values of x and y together make a square in the x-y plane.

Then plot the 2 lines x+2y=8 and x+2y=20. The area we need is the area within the square that is between the 2 lines.

It is easiest to compute the area between the 2 lines by taking the area of the square and taking away the area of the 2 triangles the lines cut off. This area between the two lines is 100-16-25=59. So the probability is 59/100

We will find the probability in terms of area enclosed . Let $O(0 , 0)$ , $A(10 , 0)$ , $B(10 ,10)$ and $C(0 ,10)$ . As $0\leq x\leq 10$ and $0\leq y\leq 10$ , the area enclosed by these two conditions is the area of $OABC$ = $100 sq. units$ . The condition $8\leq x + 2y\leq 20$ implies two enclosed areas . First , area enclosed by the line $x + 2y = 8$ , away from $O$ , since $x + 2y \geq 8$ and the lines $0\leq x\leq 10$ and $0\leq y\leq 10$ . Second , area enclosed by the line $x + 2y = 20$ , towards $O$ , since $x + 2y \leq 20$ and the lines $0\leq x\leq 10$ and $0\leq y\leq 10$ . Let the line $x + 2y = 8$ intersect $x$ -axis and $y$ -axis at $L(8 , 0)$ and $M(0 , 4)$ and let the line $x + 2y = 20$ intersect the line $x = 10$ and the line $y = 10$ at $N(10 , 5)$ and $C(0 , 10)$ . Area( $\bigtriangleup OLM$ ) = $\frac{1}{2}\times OL\times OM$ = $\frac{1}{2}\times 8\times 4$ = $16 sq. units$ . Area( $\bigtriangleup CBN$ ) = $\frac{1}{2}\times CB\times CN$ = $\frac{1}{2}\times 10\times 5$ = $25 sq. units$ Therefore , the required probability = $\frac{Area(LANCM)}{Area(OABC)}$ = $\frac{Area(OABC) - (Area(\bigtriangleup OLM) + Area(\bigtriangleup CBN))}{Area(OABC)}$ = $\frac{100 - (16 + 25)}{100}$ = $\frac{59}{100}$ = $\frac{a}{b}$ . Therefore , $a + b = 59 + 100 = 159$

Sketch the cartesian plane and let $O(0,0) A(0,10), B(10,10), C(10,0), D(20,0), E(8,0), F(0,4)$

Intersection point of straight line $AD$ and $BC$ is at $x =10$ and $x+2y=20 \Rightarrow y = 5 \Rightarrow G(10,5)$

Probability that $8 \leq x+2y \leq 20$ is

[ Area of Triangle $ADO$ $-$ Area of triangle $GCD$ $-$ Area of triangle $EOF$ ] $\div$ Area of square $ABCD$

$= [ \frac {1}{2} \times 10 \times 20 - \frac {1}{2} \times 5 \times (20 - 10) - \frac {1}{2} \times 4 \times 8 ] \div ( 10 \times 10 )$

$= \frac {59}{100}$

$a = 59, b = 100 \Rightarrow a + b = \boxed{159}$

Our domain of values are:

$0$ $\leq$ $x\leq10$ -- $(i)$

$0$ $\leq$ $y\leq10$ -- $(ii)$ which we plot on the cartesian plane.

First we find the sample area which is the solution to $(i),(ii)$ and the following inequaliy:

$0$ $\leq$ $x + 2y\leq30$ -- $(iii)$

After plotting these inequalities on the cartesian plane we find that the figure turns out to be a square with area $A=\:$ $10$ $\times$ $10$ = $100$

Now we will find the area corresponding to the required probability:

$8$ $\leq$ $x + 2y\leq20$ -- $(iv)$ which turns out to be a pentagon whose area we can calculate as :

area bounded by $(i),(ii)$ and $x + 2y\leq20$ $-$ area bounded by $(i),(ii)$ and $x + 2y\leq8$

= $\frac{1}{2}$ $\times$ $(5 + 10)$ $\times5$ $-$ $\frac{1}{2}$ $\times8$ $\times4$

= $75 - 16$

= $59=\:$ $A_{1}$ $(Say)$

So from definition of probability we have required probability = $\displaystyle{\frac{A_1}{A}}$ = $\displaystyle{\frac{59}{100}}$ = $\displaystyle{\frac{a}{b}}$ .

Comparing we obtain $a + b=\:$ $\boxed{159}.$

Consider the 10x10 square with bottom left corner at the origin in the Cartesian plane. Then the desired probability is the area of the intersection of the region $8\leq x+2y \leq 20$ with this square divided by 100. To find this area, we find the intercepts of the two boundary lines and subtract off the two triangles: $100-\frac{(10)(5)}{2}-\frac{(4)(8)}{2}=59.$ Since 59 is prime, we get 159.

Image

Consider the $Oxy$ plane and four lines : $d_1 : x=0, d_2 : x=10, d_3 : y=0, d_4 : y=10$ . The region bounded by these four lines is the square $OAFD$ with area of $100$ . Let $\ell_1,\ell_2$ be lines with equations $\ell_1 : x+2y=8, \ell_2 : x+2y=20$ . These two lines intersect with $OAFD$ at four points : $A,B,C,D,E$ . The set of $(x,y)$ satisfying the conditions $0 \le x \le 10, 0 \le y \le 10, 8 \le x+2y \le 20$ is represented by the pentagon $ABCDE$ . Therefore, the probablity is $P = \frac{S_{ABCDE}}{S_{OADE}} = 1 - \frac{S_{OBC}+S_{AFE}}{S_{OAFD}}$ . We have $B(0,4), C(8,0), E(10,5)$ , then $OB=4, OC=8, FA=10, FE=5$ . Then $S_{OBC} + S_{AFE} = \frac{1}{2} (4 \times 8 + 10 \times 5) = 41$ . Hence $P = 1 - \frac{41}{100} = \frac{59}{100}$ . The answer is $59+100 = 159$ .

On a graph, draw the lines x+2y=8 and x+2y=20.

The area enclosed between the two lines and 0<=x<=10,0<=y<=10 is the required area.

This comes out to be 59.

thus probability=59/100

The area $A_p$ of the region in the Cartesian plane defined by $0 \leq x, y \leq 10$ and $8 \leq x + 2y \leq 20$ is that of a parallelogram of height $h = 6$ and length $l = 10$ , minus the area of the triangle with vertices at $(8, 0), (10, 0), \text{and } (10, -1)$ . This triangle has an area of $1$ , so the area of our region is $59$ . The area $A_t$ of the region defined by $0 \leq x, y \leq 10$ is that of a square with side length $s = 10$ , so $A_t = 100$ . The probability that the given statements are true is thus the fraction $A_p/A_t$ = $59/100$ , which is irreducible. Thus, $a + b = 59 + 100 = \boxed{159}$ .

We can think about it graphically. The range of points (x,y) possible is a square with vertices (0,0), (0,10), (10,0), and (10,10). Now we can draw the restrictions on the graph as inequalities. Taking the first inequality, we have $8\le x+2y\implies y\ge -\dfrac{1}{2}x+4$ . This inequality, when graphed, is everything above the line that passes through the points (0,4) and (8,0) (I am describing it this way because those two points are the points on the boundary of the given range.) Now we do the same thing with the second equation. $x+2y\le 20\implies y\le -\dfrac{1}{2}x+10$ . this inequality is everything under the line that passes through the points (0,10) and (10,5). To compute the area of the resulting area (taking note of the respective ranges of x and y) we find the area of the triangle "cutoffs" and subtract that from the square. The area of the two rectangles are 25 and 16. We subtract this from the area of the square to get $100-25-16=\boxed{159}$