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Math Expert V
Joined: 02 Sep 2009
Posts: 58471
In a recent street fair students were challenged to hit one of the sha  [#permalink]

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2
4 00:00

Difficulty:   15% (low)

Question Stats: 81% (01:41) correct 19% (01:58) wrong based on 165 sessions

HideShow timer Statistics 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

Attachment: T7080.png [ 6.37 KiB | Viewed 2119 times ]

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Re: In a recent street fair students were challenged to hit one of the sha  [#permalink]

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Bunuel wrote: 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

Attachment:
T7080.png

Since, the side of each smaller equilateral = 1/3 side of larger equilateral, we can say ratio of areas = 1/9 ( as the area of equilateral triangle is sqrt(3)/4 *a^2)

Ratio of areas of 3 such smaller to the overall = 3 *1/9 = 1/3.

Hence, C
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GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: In a recent street fair students were challenged to hit one of the sha  [#permalink]

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Bunuel wrote: 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.
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Target Test Prep Representative D
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Joined: 14 Oct 2015
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Re: In a recent street fair students were challenged to hit one of the sha  [#permalink]

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Bunuel wrote: 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

Attachment:
T7080.png

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.

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Intern  B
Joined: 18 Oct 2017
Posts: 19
Location: India
Schools: ISB '21
Re: In a recent street fair students were challenged to hit one of the sha  [#permalink]

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No need of huge calculations. Since there exists only three shaded region and at a time ping pong ball can hit only one shaded region . So the probability is 1/3 ie 3 been the total of conditions elaborating as 3 shaded region. One shaded region to be hit so 1. Conclusion : 1/3 Re: In a recent street fair students were challenged to hit one of the sha   [#permalink] 08 Mar 2019, 02:58
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