Three vertices are chosen at random from the vertices of a regular hexagon then
The probability that the triangle has exactly two sides common with the side of the hexagon is

Options

A. `1/10`
B. `3/5`
C. `3/10`
D. `9/10`

Answer

To solve the problem, we need to find the probability that a triangle formed by choosing three vertices from a regular hexagon has exactly two sides that are also sides of the hexagon. Here’s the step-by-step solution:

### Step 1: Identify the total number of vertices in a regular hexagon.
A regular hexagon has 6 vertices, which we can label as A, B, C, D, E, and F.

### Step 2: Calculate the total number of ways to choose 3 vertices from the 6 vertices.
The total number of ways to choose 3 vertices from 6 is given by the combination formula:
\[ \text{Total ways} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 3: Identify the favorable outcomes where the triangle has exactly two sides common with the sides of the hexagon.
To have exactly two sides of the triangle common with the hexagon, the three chosen vertices must be consecutive vertices of the hexagon. The third vertex must not be adjacent to the two chosen vertices.

The possible combinations of vertices that form triangles with exactly two sides common to the hexagon are:
- ABC
- BCD
- CDE
- DEF
- EFA
- FAB

Thus, there are 6 favorable outcomes.

### Step 4: Calculate the probability.
The probability \( P \) that a triangle formed by choosing 3 vertices has exactly two sides common with the sides of the hexagon is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{20} = \frac{3}{10} \]

### Final Answer:
The probability that the triangle has exactly two sides common with the side of the hexagon is \( \frac{3}{10} \).

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