If three smallest squares are chosen at rando | vedclass

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(B) chessboard has $8 	imes 8 = 64$ squares. The total number of ways to choose $3$ squares out of $64$ is $\binom{64}{3} = \frac{64 	imes 63 	imes

Vedclass · Accepted Answer

$\frac{1}{434}$

Vedclass · Answer

(B) chessboard has $8 	imes 8 = 64$ squares. The total number of ways to choose $3$ squares out of $64$ is $\binom{64}{3} = \frac{64 	imes 63 	imes 62}{3 	imes 2 	imes 1} = 41664$. To have $3$ squares together in a row, in each row of $8$ squares, there are $8 - 3 + 1 = 6$ ways to choose $3$ consecutive squares. Since there are $8$ rows, the total ways for rows is $8 	imes 6 = 48$. Similarly, for columns, there are $8$ columns and $6$ ways per column, so $8 	imes 6 = 48$ ways. The total number of favorable outcomes is $48 + 48 = 96$. The probability is $\frac{96}{41664} = \frac{1}{434}$.

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

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