pentagon

[logistic_guy]

here is the question

Find the probability that a randomly chosen point in the figure lies in the shaded region.




my attemb
i can see the white area is larger so i'm certain \(\displaystyle 100\%\) the probability is less than \(\displaystyle 50\%\)
is my reason correct?

 

[Dr.Peterson]

here is the question

Find the probability that a randomly chosen point in the figure lies in the shaded region.

View attachment 38958


my attemb
i can see the white area is larger so i'm certain \(\displaystyle 100\%\) the probability is less than \(\displaystyle 50\%\)
is my reason correct?

Have you looked up the formulas for area of a triangle and of a trapezoid? Do so, and then use them.

 

[logistic_guy]

thank

Have you looked up the formulas for area of a triangle and of a trapezoid? Do so, and then use them.

the area of triangle is \(\displaystyle \frac{1}{2}bh\)
the height is the mean proportional of the base \(\displaystyle 5\) division
but the division not given

 

[Dr.Peterson]

the area of triangle is \(\displaystyle \frac{1}{2}bh\)
the height is the mean proportional of the base \(\displaystyle 5\) division
but the division not given

There's a right angle! Use it! (A base doesn't have to be horizontal ...)

And the trapezoid (trapezium to some) is easy. All the data are there, and more that you don't need.

 

[logistic_guy]

There's a right angle! Use it! (A base doesn't have to be horizontal ...)

you're righti can find the missing side and make base \(\displaystyle 3\)
i don't see this idea directly

And the trapezoid (trapezium to some) is easy. All the data are there, and more that you don't need.

i can divide it to rectangle and right triangle
but what area have to do with probability?

 

[Dr.Peterson]

you're righti can find the missing side and make base \(\displaystyle 3\)
i don't see this idea directly

A large part of doing math is finding indirect ways to solve problems, rather than doing only what you are told how to do.

i can divide it to rectangle and right triangle

Yes. Or you could use the formula for area of a trapezoid:

Trapezoid

Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

www.mathsisfun.com

but what area have to do with probability?

You implied some awareness of this in the OP:

i can see the white area is larger so i'm certain \(\displaystyle 100\%\) the probability is less than \(\displaystyle 50\%\)
is my reason correct?

A "randomly chosen point" is assumed to have probability proportional to area:

Geometric Probability | Brilliant Math & Science Wiki

Geometric probability is a tool to deal with the problem of infinite outcomes by measuring the number of outcomes geometrically, in terms of length, area, or volume. In basic probability, we usually encounter problems that are "discrete" (e.g. the outcome of a dice roll; see probability by...

brilliant.org

 

[logistic_guy]

thank Dr. very much for the resource. i like the dartboard example
i think i'm get the idea of geometrical probability

\(\displaystyle h^2 + 3^2 = 5^2\)
\(\displaystyle h^2 + 9 = 25\)
\(\displaystyle h^2 = 25 - 9 = 16\)
\(\displaystyle h = \sqrt{16} = 4\)

\(\displaystyle A_1 = \frac{1}{2}bh = \frac{1}{2}(3)(4) = 6\)

\(\displaystyle A_2 = \frac{1}{2}bh + wh = \frac{1}{2}(2)(3) + (5)(3) = 3 + 15 = 18\)

\(\displaystyle P(A_1) = \frac{A_1}{A_1 + A_2} = \frac{6}{6 + 18} = \frac{6}{24} = \frac{1}{4}\)

this mean \(\displaystyle 25\%\)

 

[Dr.Peterson]

\(\displaystyle h^2 + 3^2 = 5^2\)
\(\displaystyle h^2 + 9 = 25\)
\(\displaystyle h^2 = 25 - 9 = 16\)
\(\displaystyle h = \sqrt{16} = 4\)

\(\displaystyle A_1 = \frac{1}{2}bh = \frac{1}{2}(3)(4) = 6\)

\(\displaystyle A_2 = \frac{1}{2}bh + wh = \frac{1}{2}(2)(3) + (5)(3) = 3 + 15 = 18\)

\(\displaystyle P(A_1) = \frac{A_1}{A_1 + A_2} = \frac{6}{6 + 18} = \frac{6}{24} = \frac{1}{4}\)

this mean \(\displaystyle 25\%\)

More quickly, the triangle is the familiar 3-4-5 triangle, so its area is [imath]\frac{3\cdot4}{2}=6[/imath]; the area of the trapezoid is the product of the height and the average of the bases, [imath]3\cdot\frac{5+7}{2}=18[/imath], and the probability is the area of the triangle over the total area, [imath]\frac{6}{6+18}=25\%[/imath].

 

[logistic_guy]

More quickly, the triangle is the familiar 3-4-5 triangle, so its area is [imath]\frac{3\cdot4}{2}=6[/imath]; the area of the trapezoid is the product of the height and the average of the bases, [imath]3\cdot\frac{5+7}{2}=18[/imath], and the probability is the area of the triangle over the total area, [imath]\frac{6}{6+18}=25\%[/imath].

i appreciate your effort to write a slightly different method