JEE Advanced 1995

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Animated Solution for Mathematics - Probability: Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

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Visualized Solution

Visualizing the Regular Hexagon

Consider a regular hexagon with $6$ vertices $A, B, C, D, E, F$ .

Forming Triangles

A triangle is formed by choosing any $3$ distinct vertices out of $6$ .
Since it is a regular hexagon, no three vertices are collinear.

Total Sample Space Logic

Total ways to choose $3$ vertices from $6$ is given by combinations: $^{6} C_{3}$ .

Computing Total Triangles

$^{6} C_{3} = \frac{6 !}{3 ! ( 6 - 3 )!}$
$^{6} C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
Total possible triangles = $20$ .

Identifying Equilateral Triangles

An equilateral triangle in a regular hexagon is formed by joining alternate vertices.

First Equilateral Triangle

Choosing alternate vertices $A, C, E$ forms the first equilateral triangle $Δ A C E$ .

Second Equilateral Triangle

Choosing the remaining alternate vertices $B, D, F$ forms the second equilateral triangle $Δ B D F$ .

Setting up the Probability

Probability $P = \frac{Favorable Outcomes}{Total Outcomes}$
Favorable outcomes = $2$
Total outcomes = $20$

Final Calculation

$P = \frac{2}{20} = \frac{1}{10}$

Conclusion

The probability that the chosen triangle is equilateral is $\frac{1}{10}$ .

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The Sigma Insight: Classical Definition of Probability

Solution Diagram

Analyzing the Setup

To determine the probability of forming an equilateral triangle from the vertices of a regular hexagon, we first identify the total number of ways to select 3 vertices out of 6.

We use the combination formula:

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

Substituting our values where

n = 6

and

r = 3

^{6} C_{3} = \frac{6 !}{3 ! ( 6 - 3 )!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

Thus, the total number of possible triangles that can be formed is 20.

The Search for Equilateral Perfection

An equilateral triangle requires all sides to be equal. In a regular hexagon, this occurs only when we skip exactly one vertex between each chosen point.

Starting at vertex

A

, we skip

B

, pick

C

, skip

D

, and pick

E

. This forms our first triangle,

Δ A C E

If we start at vertex

B

, we skip

C

, pick

D

, skip

E

, and pick

F

. This forms our second triangle,

Δ B D F

If we attempt to start at vertex

C

, we skip

D

, pick

E

, skip

F

, and pick

A

, which results in the same set of vertices as

Δ A C E

. Therefore, there are exactly 2 favorable outcomes.

The Final Calculation

The probability

P

of an event is defined as the ratio of favorable outcomes to total outcomes.

P = \frac{Favorable Outcomes}{Total Outcomes} = \frac{2}{20}

Simplifying this fraction by dividing both the numerator and the denominator by 2, we obtain:

P = \frac{1}{10}

The final probability of choosing an equilateral triangle is 1/10.

Common Traps & FAQs

Why do we use combinations (

^{n} C_{r}

) instead of permutations (

^{n} P_{r}

Great question! We use combinations because the order in which we select the vertices of a triangle does not change the triangle itself. For example, choosing vertices {A, C, E} is exactly the same as choosing {C, A, E} or {E, C, A}. Since the order is irrelevant, we use combinations.

How can we be sure there are only 2 equilateral triangles?

Think about the symmetry. If you start at vertex A and skip one vertex to pick C, then skip one to pick E, you get

Δ A C E

. If you start at B and skip one to pick D, then skip one to pick F, you get

Δ B D F

. If you try to start at C, you just get the same

Δ A C E

again. There are no other ways to pick 3 vertices that are equally spaced in a regular hexagon.

What if the hexagon was not regular?

If the hexagon were not regular, the side lengths would vary, and the internal angles would not be 120 degrees. The probability would depend on the specific geometry of the hexagon, making the problem much more complex. In a regular hexagon, the symmetry is what makes the equilateral triangles possible.

Animated Solution for Mathematics - Probability: Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

Select Answer:

Visualized Solution

Visualizing the Regular Hexagon

Forming Triangles

Total Sample Space Logic

Computing Total Triangles

Identifying Equilateral Triangles

First Equilateral Triangle

Second Equilateral Triangle

Setting up the Probability

Final Calculation

Conclusion

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Search for Equilateral Perfection

The Final Calculation

Similar Questions

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Animated Solution for Mathematics - Probability: Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

Select Answer:

Visualized Solution

Visualizing the Regular Hexagon

Forming Triangles

Total Sample Space Logic

Computing Total Triangles

Identifying Equilateral Triangles

First Equilateral Triangle

Second Equilateral Triangle

Setting up the Probability

Final Calculation

Conclusion

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Search for Equilateral Perfection

The Final Calculation

Similar Questions

If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :

Three identical dice are rolled. The probability that the same number will appear on each of them is

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that a randomly selected 2-digit number belongs to the set {n∈N:(2n−2) is a multiple of 3} is equal to

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

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