probability of hitting shaded area.

[lostagain]

Here is my "problem".

Two darts are thrown at random onto the large rectangular region shown. Find the probability that both darts will land in the shaded region (small rectangle).

yikes!

Small rectangle [shaded area]
length = 3x (variable, not mult.)
width = (x + 1)

Large triangle
length = (6x + 2)
width = (2x + 2)

Any help is appreciated. Thanks.

 

[soroban]

Hello, lostagain!

The problem is ever-so-slightly vague . . .
And I think there is a typo.

Two darts are thrown at random onto the large rectangular region shown. . Of course, there is no diagramn!
Find the probability that both darts will land in the shaded region (small rectangle).

Small rectangle [shaded area]
length = (3x + 1)
width = (x + 1)

Large triangle
length = (6x + 2)
width = (2x + 2)

        * - - - - - - - - - *
        |                   |
        |   * - - - - - *   |
        |   |:::::::::::|   |
        |   |:::::::::::|   | 2x+2
        |   |:::::: x+1 |   |
        |   |:::::::::::|   |
        |   |:: 3x+1 :::|   |
        |   * - - - - - *   |
        |                   |
        * - - - - - - - - - *
                 6x+2

Area of shaded region: .\(\displaystyle (3x+1)(x+1)\)

Area of dartboard: .\(\displaystyle (6x+2)(2x+2) \:=\:2(3x+1)\cdot2(x+1) \:=\:4(3x+1)(x+1)\)

\(\displaystyle \text{Hence: }\(\text{shaded region}) \;=\;\frac{(3x+1)(x+1)}{4(3x+1)(x+1)} \;=\;\frac{1}{4}\)


\(\displaystyle \text{Therefore: }\;P(\text{both darts in shaded region}) \;=\;\frac{1}{4}\cdot\frac{1}{4} \;=\;\frac{1}{16}\)

 

[lostagain]

Sorry about the typo and thank you much for the assistance. Just what I needed. Thanks much.