Question

If three smallest squares are chosen at random on a chess board then the probability of getting them in such a way that they are all together in a row or in a column is

• A. \( \frac{73}{5208} \)
• B. \( \frac{1}{434} \)
• C. \( \frac{96}{217} \)
• D. \( \frac{479}{504} \)

Hint:

For \( k \) consecutive items in a line of \( n \) items, the number of ways is \( n - k + 1 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The problem involves calculating probability using combinations. We need to select 3 unit squares out of 64 such that they are consecutive in either a row or a column.
Step 2: Key Formula or Approach:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \] Total selections: \( \binom{64}{3} \).
Step 3: Detailed Explanation:

1. Total Outcomes:
The number of ways to choose 3 squares from 64 is: \[ n(S) = \binom{64}{3} = \frac{64 \times 63 \times 62}{3 \times 2 \times 1} = 41664 \] 2. Favorable Outcomes:
We want 3 consecutive squares. - Rows:
There are 8 rows. In one row of 8 squares, the number of groups of 3 consecutive squares is \( (8 - 3 + 1) = 6 \). Total row arrangements = \( 8 \times 6 = 48 \). - Columns:
There are 8 columns. Similarly, in one column, there are 6 groups. Total column arrangements = \( 8 \times 6 = 48 \). Total favorable cases \( n(E) = 48 + 48 = 96 \). 3. Calculate Probability:
\[ P(E) = \frac{96}{41664} \] Simplify the fraction (both divisible by 96): \[ \frac{96 \div 96}{41664 \div 96} = \frac{1}{434} \]
Step 4: Final Answer:

The probability is \( \frac{1}{434} \).