TS EAMCET · Maths · Probability
If two smallest squares are chosen at random on a chess board then the probability of getting these squares such that they do not have a side in common is
- A \(\frac{1}{18}\)
- B \(\frac{5}{36}\)
- C \(\frac{17}{18}\)
- D \(\frac{7}{36}\)
Answer & Solution
Correct Answer
(C) \(\frac{17}{18}\)
Step-by-step Solution
Detailed explanation
Total ways to choose 2 squares: \( \binom{64}{2} = \frac{64 \times 63}{2} = 2016 \) Ways to choose 2 squares with a common side: \( 8 \times 7 \text{ (horizontal)} + 8 \times 7 \text{ (vertical)} = 56 + 56 = 112 \) Ways to choose 2 squares without a common side:…
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