In a recent street fair students were challenged to hit one

[BTGmoderatorLU]

Source: Economist GMAT



In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3

The OA is C

[fskilnik@GMATH]

Source: Economist GMAT



In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region in a random point, what is the probability of hitting a shaded triangle?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3

\[? = P\left( {{\text{hit}}\,\,{\text{shaded}}\,\,{\text{region}}} \right)\]
\[\frac{{{S_{{\text{each}}\,\Delta {\text{shaded}}}}}}{{{S_{\Delta {\text{large}}}}}} = {\left( {\frac{1}{3}} \right)^2} = \frac{1}{9}\,\,\,\,\,\,\,\left[ {\,{\text{each}}\,\,\Delta {\text{shaded}}\,\,{\text{is}}\,\,{\text{similar}}\,\,{\text{to}}\,\,{\text{the}}\,\,\Delta {\text{large}}\,} \right]\]
\[? = 3 \cdot \frac{1}{9} = \frac{1}{3}\,\,\,\,\,\,\left( {{\text{geometric}}\,\,{\text{probability}}} \right)\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

[Scott@TargetTestPrep]

Source: Economist GMAT



In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3

We see that each of the smaller shaded equilateral triangle has the same area. Furthermore, the unshaded region is a regular hexagon that can divide into 6 equilateral triangles each equalling to the area of a shaded triangle. Thus there are 3 + 6 = 9 equilateral triangles of the same area and the probability hitting a shaded triangle is 3/9 = 1/3.

Alternate Solution:

Let's assume that each side of the large triangle is 6 units. The area of the large triangle is thus (1/2)(6)(6√3) = 18√3.

A side of any of the shaded triangles is 2. The area of one shaded triangle is (1/2)(2)(2√3) = 2√3. There are 3 shaded triangles, so their total area is 6√3.

The probability of hitting any shaded triangle is the total area of the shaded triangles divided by the total area of the entire large triangle: 6√3 / 18√3 = 1/3.

Answer: C