If three of the six vertices of a regular hexagon are chosen at random, then the probability that th

Q: Wait, why do we use $^6C_3$ and not $6 \times 5 \times 4$?

Great question! We use $^6C_3$ because the order in which we select the vertices does not matter. Whether you pick vertex $A_1$, then $A_3$, then $A_5$, or $A_5$, then $A_1$, then $A_3$, you are still forming the exact same triangle $\triangle A_1A_3A_5$. Since the order is irrelevant, we use combinations, not permutations.

Q: How can I be sure there are only two equilateral triangles?

Think about the symmetry of the hexagon. If you pick a vertex, you must skip exactly one vertex to maintain the $120^\circ$ angle required for an equilateral triangle. If you start at $A_1$, you are forced to pick $A_3$ and $A_5$. If you start at $A_2$, you are forced to pick $A_4$ and $A_6$. If you try to start at $A_3$, you end up with the set $\{A_3, A_5, A_1\}$, which is the same as the first one. Thus, there are only two distinct sets.

Q: Can I solve this using coordinate geometry instead?

You absolutely could! You could assign coordinates to the vertices of a regular hexagon (e.g., using complex numbers or trigonometry) and check the distance between every pair of points for all 20 combinations. However, that is a much longer path. In competitive exams like JEE, geometric intuition is your best friend—it saves time and reduces calculation errors.

Question

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 :

Options:

(A) $3/10$

(B) $1/10$

(C) $3/20$

(D) $1/5$

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: A Hexagonal Journey

Imagine you are standing before a perfect, regular hexagon. It is a shape of profound balance, a staple of nature's most efficient designs. Today, we are not just solving a probability problem; we are exploring the hidden symmetry within this six-sided wonder.

Our task is simple yet elegant: if we randomly choose three vertices from the six available, what is the probability that they form an equilateral triangle? Let us break this down, step by step.

### Phase 1: The Total Sample Space

Before we can find the probability, we must understand the scope of our possibilities. We have six distinct vertices, which we can label $A_1, A_2, A_3, A_4, A_5, A_6$.

We need to select exactly three of these to form a triangle. Because the triangle $	riangle A_1A_3A_5$ is identical to $	riangle A_5A_1A_3$, the order of selection does not matter. This is the classic definition of a combination. We use the formula $^nC_r$, where $n=6$ and $r=3$:

$$n(S) = ^6C_3 = \frac{6!}{3!(6-3)!}$$

Expanding this, we get $\frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1}$. The $3 	imes 2 	imes 1$ in the denominator is $6$, which cancels out the $6$ in the numerator.

We are left with $5 	imes 4$, which equals $20$. So, there are exactly $20$ possible triangles we can form from these six vertices. This is our total sample space.

### Phase 2: The Geometric Insight

Now, we move from the algebraic to the geometric. What makes a triangle equilateral in a regular hexagon? If you visualize the hexagon, you will notice that the internal angle is $120^\circ$.

To create an equilateral triangle, the vertices must be spaced perfectly. If you pick a vertex, you must skip exactly one vertex to reach the next one. This creates a triangle where each side spans two edges of the hexagon, ensuring all sides are equal.

Let us trace this. If we start at $A_1$, we skip $A_2$ and pick $A_3$. Then we skip $A_4$ and pick $A_5$. Connecting $A_1, A_3,$ and $A_5$ gives us our first equilateral triangle: $	riangle A_1A_3A_5$.

Now, what if we start at $A_2$? We skip $A_3$ and pick $A_4$. Then we skip $A_5$ and pick $A_6$. Connecting $A_2, A_4,$ and $A_6$ gives us our second equilateral triangle: $	riangle A_2A_4A_6$.

Are there any more? If we try to start at $A_3$, we would pick $A_5$ and $A_1$, which is just the first triangle again. We have exhausted all unique possibilities. Thus, the number of favorable outcomes, $n(E)$, is exactly $2$.

### Phase 3: The Final Calculation

We have reached the home stretch. The probability $P(E)$ of an event is simply the ratio of favorable outcomes to the total sample space:

$$P(E) = \frac{n(E)}{n(S)}$$

Substituting our values, we get:

$$P(E) = \frac{2}{20}$$

Simplifying this fraction by dividing both the numerator and the denominator by $2$, we arrive at our final result:

$$P(E) = \frac{1}{10}$$

There we have it! A one in ten chance. It is a wonderful reminder that even in complex problems, the answer often lies in visualizing the underlying geometry. **The final probability is $\frac{1}{10}$.**

Animated Solution for Mathematics - Probability: 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 :

Select Answer:

Visualized Solution

The Regular Hexagon

Total Sample Space

Setting up $n (S)$

Calculating $n (S)$

Condition for Equilateral Triangles

The First Equilateral Triangle

The Second Equilateral Triangle

Total Favorable Outcomes

Probability Formula

Final Calculation

Conclusion

The Sigma Insight: Classical Definition of Probability

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Similar Questions

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

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

An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that, among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list?

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

The probability that a randomly selected 2-digit number belongs to the set ${n \in N : (2^{n} - 2) is a multiple of 3}$ is equal to

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

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls 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