4. Two randomly selected squares are to be removed from a chess board, if every pair to be
removed is equally likely, the probability that the two removed squares are neither in the
same row, nor in the same column, nor in the same diagonal-main or off is
(a) less than 1/3
(b) more than 1/3, but, less than 2/3
(c) more than 2/3, but, less than 3/4
(d) more than 3/4
Answers
Answer:
Clearly, the first square can be chosen in (64C1) = 64 ways.
For every choice of the first square, total 15 squares are not available for the choice of second square (7 other squares lengthwise, 7 other squares breadthwise and 1 square for that one already chosen as the first square). So, for every choice of the first square, (49C1) = 49 ways one can choose the second square.
But, one more knot is left to be untied!
As per above procedure, the entire selection process is doubled. This is because, suppose, in any outcome, we selected the square at (1, 1) as the first square and (2, 3) as the second square. Now, in another outcome, the squares at (2, 3) and (1, 1) may be the first and the second choice respectively. But, these two outcomes lead to the same duo; hence, these two outcomes give same result.
So, the answer to this question is [(64*49) / 2] = 1568.
Step-by-step explanation:
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Answer:
d
Step-by-step explanation: