Question

If two of the 64 squares are chosen at random on a chessboard, the probability that they have a side in common is
(a)$\dfrac{1}{9}$
(b)$\dfrac{1}{18}$
(c) $\dfrac{2}{7}$
(d) None of these

Answer 1

To solve this question, firstly we will divide the solution into 3 cases, where case A will be when we consider 4 corner square, case B will be when we consider side square apart from corner square, and case C will be when we consider inner square. Then for the probability of an event, that two square have a common side, we will add probabilities of each case.

Complete step-by-step solution:
Let categorize the cases for the given condition in question.
Let Case A denotes that we choose one of the four corners of the chessboard, then we have two squares for each corner who have a common side with a corner square.
So, P(A) $=\dfrac{^{4}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{2}{{C}_{1}}}{^{63}{{C}_{1}}}$, this because as we have selected one corner from total four corners and 1 out of total 64 squares that is why $\dfrac{^{4}{{C}_{1}}}{^{64}{{C}_{1}}}$ and for each corner we have 2 possible squares having common side and we are limited to take two squares at a time so, we choose 1 square out of 2 squares and also which is 1 out of 63 squares as we have already chosen one corner that is why $\dfrac{^{2}{{C}_{1}}}{^{63}{{C}_{1}}}$.
Let Case B denotes that we choose any side square apart from all the four corners of the chessboard, then we have 24 such squares which are on the side apart from the corner square, and for each side square, we have 3 squares who have common sides with any of the side squares apart from a corner.
So, P ( B ) $=\dfrac{^{24}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{3}{{C}_{1}}}{^{63}{{C}_{1}}}$, this because as we have selected any side square apart from all the four corners of the chessboard and same is 1 out of total 64 squares that is why $\dfrac{^{24}{{C}_{1}}}{^{64}{{C}_{1}}}$ and for any side square apart from all the four corners of the chessboard, we have 3 possible squares having common side and we are limited to take two squares at a time so, we choose 1 square out of 2 squares and also which is 1 out of 63 squares as we have already chosen one corner that is why $\dfrac{^{3}{{C}_{1}}}{^{63}{{C}_{1}}}$.
Let, case C denotes inner squares, so for each 6 columns we have 6 inner squares, then we have a total 36 inner square, then we have 4 such squares which are sharing common sides with any of the one selected square.
So, P(C) $=\dfrac{^{36}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{4}{{C}_{1}}}{^{63}{{C}_{1}}}$, this is because as we have selected any inner square and same is 1 out of total 64 squares that is why $\dfrac{^{36}{{C}_{1}}}{^{64}{{C}_{1}}}$ and for any inner square, we have 4 possible squares having common side and we are limited to take two squares at a time so, we choose 1 square out of 2 squares and also which is 1 out of 63 squares as we have already chosen one corner that is why $\dfrac{^{3}{{C}_{1}}}{^{63}{{C}_{1}}}$.
Let, an event that is two square having a common side, be denoted by X, then
So, total P( X ) = P ( A ) + P ( B ) + P ( C ),
So, P ( X ) $=\dfrac{^{4}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{2}{{C}_{1}}}{^{63}{{C}_{1}}}+\dfrac{^{24}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{3}{{C}_{1}}}{^{63}{{C}_{1}}}+\dfrac{^{36}{{C}_{1}}}{^{64}{{C}_{1}}}\times \dfrac{^{4}{{C}_{1}}}{^{63}{{C}_{1}}}$
We know that, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $^{n}{{C}_{1}}=n$, so
So, using the above formula and property, we get
P ( X ) $=\dfrac{4}{64}\times \dfrac{2}{63}+\dfrac{24}{64}\times \dfrac{3}{63}+\dfrac{36}{64}\times \dfrac{4}{63}$
On simplifying, we get
P ( X ) $=\dfrac{8}{64\times 63}+\dfrac{72}{64\times 63}+\dfrac{144}{64\times 63}$
On solving, we get
P ( X ) $=\dfrac{8+72+144}{64\times 63}$
P ( X ) $=\dfrac{224}{64\times 63}$
P ( X ) $=\dfrac{1}{18}$
Hence, option ( b ) is the correct answer.

Note: To solve such questions, one must consider each and every case carefully as if any case is not considered, then the answer will get wrong. Remember that chess has a total of 8 columns and 8 rows hence, 64 squares. Try not to make any mistakes in calculation.