Chapter A.4, Problem 76E

To determine

Find the probability that marble will come to rest in shaded portion of the base

Answer to Problem 76E

The probability is $\frac{x^2}{8{(x+2)}^{}}$ where, $x\ne 0$

Explanation of Solution

Given:

The probability that the marble will come to rest in the shaded portion of the base is equal to the ratio of the shaded area to the total area of the figure.

  

Consider the above figure

In which the sides of the bigger triangle is $x+4$ and $x$ respectively , and the bases are $x+2+\frac{4}{x}(x+2)$ and $x+2$ respectively.

Also consider that,

The area of the bigger triangle

  $\begin{array}{l}=\frac{1}{2}(x+4)\left(x+2+\frac{4}{x}(x+2)\right)\hfill \\ =\frac{(x+4)(x+2)\left(1+\frac{4}{x}\right)}{2}\hfill \\ =\frac{{(x+4)}^2(x+2)}{2x}\hfill \end{array}$

Also, the area of the smaller triangle $=\frac{1}{2}(x)(x+2)$ .

Now area of the shaded region is area of bigger triangle minus the area of smaller triangle.

Therefore area of shaded region is

  $\begin{array}{l}=\frac{{(x+4)}^2(x+2)}{2x}-\frac{1}{2}(x)(x+2)\hfill \\ =\frac{x+2}{2}\left(\frac{{(x+4)}^2}{x}-x\right)\hfill \\ =\frac{x+2}{2}\left(\frac{x^2+16+8x-x^2}{x}\right)\hfill \\ =\frac{8{(x+2)}^2}{2x}\hfill \\ =\frac{4{(x+2)}^2}{x}\hfill \end{array}$

Given that the probability that the marble tossed will end up at rest in the shaded portion is equal to the ratio of the area of shaded region to the area of the total figure.

Hence, the probability is

  $=\frac{\frac{1}{2}x(x+2)}{\frac{4{(x+2)}^2}{x}}$

  $=\frac{x^2}{8{(x+2)}^{}}$ where, $x\ne 0$