A board has 16 squares (a $4 imes4$ grid). Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

Question

A board has 16 squares (a $4	imes4$ grid). Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Total pairs $\binom{16}2=120$. Pairs sharing a side: horizontal $4	imes3=12$ and vertical $12$, so $24$. Probability $=\frac{120-24}{120}=\frac45$. (Medium confidence: the printed figure shows a grid that looks $4	imes5$, but the question states 16 squares and the key $\frac45$ corresponds to a $4	imes4$ board.) Answer (2).

Q. A board has 16 squares (a $4\times4$ grid). Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

Explanation

Reviewed by Dr. Abubacker

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