Three squares of Chess board are selected at random. Find the probability of getting 2 squares of one colour and other of a different colour.

Options

A. `13/21`
B. `14/21`
C. `8/21`
D. `16/21`

Explanation

To solve the problem of finding the probability of selecting 3 squares from a chessboard such that 2 squares are of one color and 1 square is of a different color, we can follow these steps:

### Step 1: Determine the number of squares of each color on the chessboard.
A standard chessboard has 64 squares in total, with 32 squares being black and 32 squares being white.

**Hint:** Remember that a chessboard has an equal number of black and white squares.

### Step 2: Identify the possible combinations of squares.
We need to select either:
- 2 white squares and 1 black square, or
- 2 black squares and 1 white square.

### Step 3: Calculate the number of ways to choose the squares for each combination.
1. **For 2 white squares and 1 black square:**
- The number of ways to choose 2 white squares from 32 white squares is given by the combination formula \( \binom{n}{r} \):
\[ \text{Ways to choose 2 white squares} = \binom{32}{2} \]
- The number of ways to choose 1 black square from 32 black squares:
\[ \text{Ways to choose 1 black square} = \binom{32}{1} \]

2. **For 2 black squares and 1 white square:**
- The number of ways to choose 2 black squares from 32 black squares:
\[ \text{Ways to choose 2 black squares} = \binom{32}{2} \]
- The number of ways to choose 1 white square from 32 white squares:
\[ \text{Ways to choose 1 white square} = \binom{32}{1} \]

### Step 4: Combine the two cases.
The total number of favorable outcomes is the sum of the two cases:
\[ \text{Total favorable outcomes} = \binom{32}{2} \cdot \binom{32}{1} + \binom{32}{2} \cdot \binom{32}{1} = 2 \cdot \binom{32}{2} \cdot \binom{32}{1} \]

### Step 5: Calculate the total number of ways to choose any 3 squares from 64 squares.
The total number of ways to choose 3 squares from 64 squares is:
\[ \text{Total ways to choose 3 squares} = \binom{64}{3} \]

### Step 6: Calculate the probability.
The probability \( P \) of selecting 2 squares of one color and 1 square of a different color is given by:
\[ P = \frac{\text{Total favorable outcomes}}{\text{Total ways to choose 3 squares}} = \frac{2 \cdot \binom{32}{2} \cdot \binom{32}{1}}{\binom{64}{3}} \]

### Step 7: Substitute the values and simplify.
Calculating the combinations:
- \( \binom{32}{2} = \frac{32 \cdot 31}{2} = 496 \)
- \( \binom{32}{1} = 32 \)
- \( \binom{64}{3} = \frac{64 \cdot 63 \cdot 62}{3 \cdot 2 \cdot 1} = 39711 \)

Substituting these values:
\[ P = \frac{2 \cdot 496 \cdot 32}{39711} = \frac{31616}{39711} \]

### Step 8: Final simplification.
After simplifying \( \frac{31616}{39711} \), we find that it reduces to \( \frac{16}{21} \).

Thus, the required probability is:
\[ \boxed{\frac{16}{21}} \]